Class 11 Physics

Unit 1 Physical World And Measurement Chapter 1 Measurement And Dimension Of Physical Quantity Multiple Choice Questions

Question 1. X, Y, and Z are three physical quantities having units g · cm² · s^-5, g · s^-1, and cm · s^-2. Relationship between X, Y, and Z can be represented by

1. X ∝  YZ
2. X ∝ YZ²
3. X ∝ Y²Z
4. X ∝ Y/Z

Answer: 2. X ∝ YZ²

Question 2. Steradian is the unit of

1. Angle
2. Solid angle
3. Arc of a circle
4. Circumference

Answer: 2. Solid angle

Question 3. The dimension of electric potential is the same as that of

1. Current
2. Force
3. Electromotive force
4. Energy

Answer: 3. Electromotive force

Question 4. The dimensional formula of potential difference is

1. ML²T^-3A^-1
2. MLT^-3A^-1
3. ML³T^-3A
4. ML²T^-3A^-2

Answer: 1. ML²T^-3A^-1

Question 5. Which of the following is dimensionally correct?

1. Pressure = energy per unit volume
2. Pressure = energy per unit area
3. Pressure = force per unit length
4. Pressure = force per unit volume

Answer: 1. Pressure = energy per unit volume

Question 6. The unit of specific conductivity is

1. Ω • cm^-1
2. Ω^-1 • cm^-1
3. Ω • cm^-2
4. Ω^-1 • cm

Answer: 2. Ω^-1 • cm^-1

Question 7. As per quantum theory, the energy E of a photon is related to its frequency v as E = hv, where hv = Planck’s constant. Then, the dimension of h would be

1. ML²T^-2
2. ML²T^-1
3. MLT^-2
4. MLT^-1

Answer: 2. ML²T^-1

Question 8. Which of the following quantities is dimensionless, if v = speed, r = radius of a circular path, and g = acceleration due to gravity?

1. \(\frac{v^2 r}{g}\)
2. \(\frac{v^2}{r g}\)
3. \(\frac{v^2 g}{r}\)
4. \(v^2 r g\)

Answer: 2. \(\frac{v^2}{r g}\)

Question 9. Considering force (F), length (L), and time (T) to be fundamental physical quantities, find the dimension of mass.

1. FL^-1T²
2. FL^-1T^-1
3. FLT^-2
4. F^-1L^-1T²

Answer: 1. FL^-1T²

Question 10. The time period of a pendulum (T), its length (l), the mass of its bob (m), and acceleration due to gravity (g) are related as T = km^xl^yg^z where,

1. x = 1, y = 1/2, z = 1
2. x = 0, y =1/2 z = 1
3. x = 1, y = z = 1/2
4. 0, y = 1/2 z = -1/2

Answer: 4. 0, y = 1/2 z = -1/2

Question 11. The equation of the state of a gas is given by \(\left(p+\frac{a}{V^3}\right)\left(V-b^2\right)=c T\), where p, V, and T are pressure, volume and temperature respectively and a, b, c are constants. The dimensions of a and b are respectively

1. \(\mathrm{ML}^8 \mathrm{~T}^{-2} and \mathrm{L}^{3 / 2}\)
2. \(\mathrm{ML}^5 \mathrm{~T}^{-2} and \mathrm{L}^3\)
3. \(\mathrm{ML}^5 \mathrm{~T}^{-2} and \mathrm{L}^6\)
4. \(\mathrm{ML}^6 \mathrm{~T}^{-2} and \mathrm{L}^{3 / 2}\)

Answer: 1. \(\mathrm{ML}^8 \mathrm{~T}^{-2} and \mathrm{L}^{3 / 2}\)

Question 12. The unit of t is s and that of x is m in the expression, y = Acos(\(\frac{t}{p}\) – qx). Then

1. x and q have the same unit
2. x and p have the same unit
3. t and q have the same unit
4. t and p have the same unit

Answer: 4. t and p have the same unit

Question 13. The unit of both x and y is m in the expression y = \(A \sin \left[\frac{2 \pi}{\lambda}(c t-x)\right]\). Then

1. x,  and λ have the same unit
2. x and λ have the same unit, but the unit of A is different
3. c and \(\frac{2\pi}{\lambda}\) have the same unit
4. (ct-x) and \(\frac{2\pi}{\lambda}\) have the same unit

Answer: 1. x,  and λ have the same unit

Question 14. The dimension of ω in the expression y = Asin(ωt- kx) is

1. M⁰LT
2. M⁰L^-1T⁰
3. M⁰L⁰T^-1
4. M⁰LT^-1

Answer: 3. M⁰L⁰T^-1

Question 15. If the error in the measurement of the momentum of a particle is 50% then the error in the measurement of kinetic energy is

1. 75%
2. 100%
3. 125%
4. 200%

Answer: 2. 100%

Question 16. Percentage errors in the measurement of mass and speed are 2% and 3% respectively. The error in the estimation of kinetic energy obtained by measuring mass and speed will be

1. 12%
2. 8%
3. 10%
4. 2%

Answer: 2. 8%

Question 17. Choose the incorrect statement out of the following.

1. Every measurement by any measuring instrument has some errors
2. Every calculated physical quantity based on measured values has some error
3. A measurement can have more accuracy but less precision and vice versa
4. The percentage error is different from the relative error

Answer: 4. The percentage error is different from the relative error

Question 18. A screw gauge has a pitch of 0.5 mm and has 50 divisions on its circular scale. Its linear and circular scale readings are, respectively, 2 mm and 2 s when the diameter of a sphere is measured by the gauge. The result of this measurement is

1. 1.2 mm
2. 1.25 mm
3. 2.20 mm
4. 2.25 mm

Answer: 4. 2.25 mm

Question 19. The smallest main scale division of a vernier caliper = 1 mm. Here, 20 vernier divisions =16 main scale divisions. The vernier constant is

1. 0.02 mm
2. 0.05 mm
3. 0.1mm
4. 0.2 mm

Answer: 4. 0.2 mm

Question 20. The measured values of mass, radius and length of a wire are, respectively, (0.3 ± 0.003) g, (0.5 ± 0.005) mm, and (6 ± 0.06) cm. The maximum percentage error in the computed value of the density of the material of the wire would be

1. 1
2. 2
3. 3
4. 4

Hint: Fractional errors : \(\frac{\Delta m}{m}=\frac{0.003}{0.3}=\frac{1}{100}\);

Similarly, \(\frac{\Delta r}{r}=\frac{\Delta l}{l}=\frac{1}{100}\)

Density \((D)=\frac{m}{\pi r^2 l}\);

So, \(\frac{\Delta D}{D}=\frac{\Delta m}{m}+2 \frac{\Delta r}{r}+\frac{\Delta l}{l}=\frac{1}{100}+\frac{2}{100}+\frac{1}{100}=\frac{4}{100}=4 \%\)

Answer: 4. 4

Question 21. The dimension of a quantity is M^aL^bT^-c. If the measurements of mass, length, and time involve errors of α%, β%, and γ%, respectively, then the error in the measurement of the given quantity is

1. (αa-βb+γc)%
2. (αa+ βb + γc)%
3. (αa +βb-γc)%
4. (αa-βb-γc)%

Answer: 2. (αa+ βb + γc)%

Question 22. What is the number of significant figures in (3.10 + 4.60) x 10⁷?

1. 5
2. 3
3. 4
4. 7

Answer: 2. 3

Question 23. The sum of 6.75 x 10³ cm and 4.52 x 10² cm, keeping the significant digits only, is

1. 7.202 x 10³ cm
2. 72.0 x 10² cm
3. 0.72 x 10⁴ cm
4. 7.20 x 10² cm

Answer: 2. 72.0 x 10² cm

Question 24. A rectangle has a length = 4.234 m and breadth = 1.05 m. What is its area in m²? Keep the significant digits only.

1. 4.4457
2. 4.45
3. 4.446
4. 4.44

Answer: 2. 4.45

In these type of questions, more than one option are correct.

Question 25. Photon is the quantum of radiation with energy E = hv where v is the frequency and h is Planck’s constant. The dimension of h is the same as that of

1. Linear impulse
2. Angular impulse
3. Linear momentum
4. Angular momentum

Answer:

2. Angular impulse

4. Angular momentum

Question 26. Choose the correct statements:

1. If a physical equation is dimensionally correct, it always expresses the correct relation among the quantities involved.
2. Even if a physical equation is dimensionally correct, it may not express the correct relation among the quantities involved.
3. If a physical equation is dimensionally incorrect, cannot express the correct relation among the quantities involved.
4. Even if a physical equation is dimensionally incorrect, it may express the correct relation among the quantities involved.

Answer:

2. Even if a physical equation is dimensionally correct, it may not express the correct relation among the quantities involved

3. If a physical equation is dimensionally incorrect, cannot express the correct relation among the quantities involved.

Question 27. A wave progresses along the x-axis with time t. The instantaneous displacement from the mean position is given by. y = asin[(ωt- kx) + θ]. Then, the dimensions are,

1. [ω] = T^-1
2. [a] = L
3. [θ] = 1
4. [k] = L^-1

Answer:

Question 28. The dimension of pressure is equal to that of

1. The force exerted per unit area
2. Energy density
3. Change in momentum per unit area per second
4. Momentum per unit volume

Answer:

1. The force exerted per unit area
2. Energy density
3. Change in momentum per unit area per second
4. Momentum per unit volume

Question 29. If P, Q, and R are physical quantities having different dimensions, which of the following can never be a meaningful quantity?

1. (P-Q)/R
2. PQ-R
3. PQ/R
4. (R+Q)/P

Answer:

1. (P-Q)/R

4. (R+Q)/P

Properties Of Bulk Matter – Change Of State Of Matter Introduction

Matter generally exists in three states or phases—solid, liquid and gaseous. When a substance changes from one state to another, it is said to have undergone a change of state or phase change or phase transition.

As matter changes from the solid to its liquid state, the phenomenon is called melting or fusion. The reverse phenomenon i.e., change from the liquid to the solid state is termed as freezing or solidification.

• Similarly, the change of a liquid to a gas, and the change from a gaseous state to the liquid state are called vaporisation and condensation respectively. Sublimation is the direct conversion from solid to gaseous state.
• To bring about a change of state of a substance, the application of heat to or the extraction of heat from the substance is essential. The change of state of a substance, during which heat is absorbed by the substance is called a higher change of state (for example: melting and vaporisation).
• On the other hand, the change of state of a substance, during which heat is extracted from the substance is called lower change of state (for example: solidification and condensation.

Read and Learn More: Class 11 Physics Notes

Properties Of Bulk Matter – Change Of State Of Matter Latent Heat

When water boils at 100°C it does not change its temperature until it completely converts into vapour. So the applied heat changes the water from its liquid state to its vapour state.

• It, too, cannot be measured. Again, the temperature of a mass of at 100°C does not fall until the vapour completely converts into water by releasing heat.
• Latent Heat Definition: The amount of heat extracted or applied to change the state of unit mass of a substance at a constant temperature, is called the latent heat of the substance for that change of state.

Hence, corresponding to different changes of state, latent heat are of four types:

1. Latent heat of fusion
2. Latent heat of solidification
3. Latent heat of vaporisation
4. Latent heat of condensation

Explanation of latent heat: The molecules of a solid substance are arranged in a fixed crystalline structure and therefore the solid has a definite shape. On the other hand, the molecules of a liquid are not so arranged and so they have no fixed shape.

• Hence, change of a solid to its liquid state, or melting, means breaking up of the crystalline structure. Latent heat supplies the necessary energy for breaking this crystalline arrangement of molecules and hence there is no change in temperature.
• Similarly, at atmospheric pressure, the intermolecular attraction in gases is very small in comparison to that in solids and in liquids. The intermolecular separations in gases are much higher.
• Hence, change of a liquid to its gaseous state, i.e., vaporisation, means the total separation of the molecules from their mutual attraction. The energy required for this purpose is supplied by the latent heat.

Since the energy required to increase the intermolecular separation in changing a liquid to gas is much higher than that in changing a solid to liquid latent heat of vaporisation is higher than the latent heat of fusion for a substance.

Properties Of Bulk Matter – Change Of State Of Matter Triple Point

There is a pressure and temperature at which the solid, liquid and gaseous states of a substance can coexist in equilibrium. This is called the triple point of that substance.

The graph of pressure (p) versus temperature (T) for a substance is sometimes known as phase diagram. the phase diagrams of water and carbon dioxide respectively. Phase diagram generally divides the p- T plane in three regions—solid, liquid and gaseous regions.

Lines BO, AO and CO are called sublimation curve, fusion curve and vaporisation curve respectively. The points on the sublimation curve denote the pressures and temperatures at which the solid and gaseous states of the substance coexist in equilibrium.

Similarly, the points on fusion curve denote pressures and temperatures at which the solid and liquid states coexist in equilibrium. Again, the points on vaporisation curve denote pressures and temperatures at which the liquid and gaseous states coexist in equilibrium.

The curves BO, AO and CO meet at point O. The point O denotes the pressure and temperature at which the three states—solid, liquid and gaseous states— coexist in equilibrium. This means that the point O is the triple point. The pressure and temperature at triple point of water are 4.58 mm Hg and 0.01°C respectively.

Properties Of Bulk Matter – Change Of State Of Matter Triple Point Numerical Examples

Example 1. A copper calorimeter of mass 100 g contains a mixture of ice and water of total mass 40 g. A piece of copper of mass 100 g at 100°C is dropped in the calorimeter and finally, the temperature of the system becomes 10°C. Find the mass of ice in the initial ice-water mixture. Specific heat of copper = 0.09 cal · g^-1 · °C^-1 and latent heat of fusion of ice = 80 cal · g^-1.
Solution:

A copper calorimeter of mass 100 g contains a mixture of ice and water of total mass 40 g. A piece of copper of mass 100 g at 100°C is dropped in the calorimeter and finally, the temperature of the system becomes 10°C.

Let initial mass of ice in the calorimeter = xg.

Hence, mass of water initially present at 0°C was (40 – x) g.

Heat lost by the copper piece

= 100 x 0.09 x (100 – 10) = 810 cal

Heat gained by ice for melting = x x 80 = 80x cal and heat gained by that melted ice to raise its temperature to 10°C = xx 1 x10 cal = 10 xcal.

Heat gained by (40 – x) g of water to come up to 10°C = (40 – x) x 1 x 10 cal.

Heat taken by the calorimeter = 100 x 0.09 x 10 = 90 cal

As heat lost = heat gained,

80x+ 10x+ (40 -x) x 10 + 90 = 810 or, 80x = 320

∴  x = 4

∴ There was 4 g of ice in the mixture.

Example 2. A calorimeter of water equivalent 50 g contains 250 g of water and 200 g of ice at 0°C. 200 g of steam at 100°C is passed through the ice-water mixture. Find the final temperature of the mixture and its total mass in the calorimeter. Latent heat of fusion of ice = 80 cal · g^-1 and that of vaporisation of water = 540 cal · g^-1.
Solution:

A calorimeter of water equivalent 50 g contains 250 g of water and 200 g of ice at 0°C. 200 g of steam at 100°C is passed through the ice-water mixture.

Heat given up due to condensation of 200 g of steam at 100°C

= 200 x 540 = 108000 cal

Heat absorbed by 200 g ice at 0°C to form water at 0°C = 200 x 80 = 16000 cal.

Heat absorbed by the calorimeter, water and melted ice to reach 100°C from 0°C = (50 + 250 + 200) x 100 = 50000 cal

Maximum heat that could be taken in =16000 + 50000 = 66000 cal, which is much less than 108000 cal.

1. The final temperature of mixture is 100°C and the entire 200 g of steam is not condensed.

This mixture contains molten ice, water and x g (say) of condensed steam. Hence, (200 -x)g of steam passes through as steam.

2. Heat given up = x x 540 cal and heat taken in = 66000 cal.

∴ x x 540 = 66000

∴ x = \(\frac{66000}{540}=122.2 \mathrm{~g}\)

Total mass of the mixture = water + molten ice + con¬densed steam =250 + 200 + 122.2 = 572.2 g.

Example 3. A metal piece of mass 48.5 g at 10.7°C, placed in a flow of steam can condense 0.762 g of steam. Find the specific heat capacity of the metal. Latent heat of steam = 537 cal · g^-1.
Solution:

A metal piece of mass 48.5 g at 10.7°C, placed in a flow of steam can condense 0.762 g of steam.

Let the specific heat capacity of the metal be s cal · g^-1 · °C^-1.

Due to the flow of steam, the temperature of the metal rises to 100°C on condensation of 0.762 g of steam.

Hence, heat given up by the steam = 0.762 x 537 cal.

Heat received by the metal =48.5 x s x (100 – 10.7) cal

∴ Heat lost = heat gained

∴ 0.762×537 = 48.5 xs(100-10.7)

or, s = 0.094

Therefore, the specific heat capacity of the metal is 0.094 cal · g^-1 · °C^-1.

Example 4. In an Industrial unit, 10 kg of water is to be heated per hour from 20°C to 80°C. To achieve this, steam from a boiler at 150°C is circulated through a copper coil immersed in water. Steam condenses inside the coil, changes to water at 90°C and flows back to the boiler for recirculation. Estimate the mass of steam in kg, required per hour. Specific heat of steam = 1 cal · g^-1 · °C^-1 and latent heat of steam = 540 cal · g^-1.
Solution:

In an Industrial unit, 10 kg of water is to be heated per hour from 20°C to 80°C. To achieve this, steam from a boiler at 150°C is circulated through a copper coil immersed in water. Steam condenses inside the coil, changes to water at 90°C and flows back to the boiler for recirculation.

Amount of heat required per hour by 10 kg of water = 10 x 10³ x 1 x (80 – 20) = 60 x 10⁴ cal.

Let the mass of steam required be x kg.

Hence, heat given up by the steam

= x x 10³ x 1 x (150- 100) + x x 10³ x 540 + x x 10³ x 1 x (100 – 90)

= x x 10³ x (50 + 540 + 10) = x x 10⁴ x 60 cal

According to the problem, x x 10⁴ x 60 = 60 x 10⁴

or x = 1

Thus the mass of tne steam requires per hour is 1 kg.

Example 5. 10 g of a solid at -10°C needs 64 cal of heat to rise up to -2°C remaining in its solid state. To raise its temperature from -10°C to 1°C and 3°C (in liquid state), required heat are 880 cal and 900 cal respectively. If the specific heats of the material in its solid and liquid states be s₁ and s₂, calculate their values. If L is the latent heat of fusion and t_m is the melting point in °C, then show that, L = 79 + 0.2 t_m.
Solution:

10 g of a solid at -10°C needs 64 cal of heat to rise up to -2°C remaining in its solid state. To raise its temperature from -10°C to 1°C and 3°C (in liquid state), required heat are 880 cal and 900 cal respectively. If the specific heats of the material in its solid and liquid states be s₁ and s₂

Heat required to change the temperature of the solid from -10°C to -2°C = 10s₁{-2-(-10)} = 80s₁.

According to the problem, 80s₁ = 64.

∴ s₁ = 0.8 cal · g^-1 · °C^-1

Difference in the heat required to raise the temperature of the object in its liquid state from -10°C to 1°C and from -10°C to 3°C = 10 x s₂ x (3 – 1) = 20s₂.

According to the problem, 20s₂ = 900 – 880 = 20 or, s₂ = 1 cal · g^-1 · °C^-1

If the melting point is t_m, heat required to change the body in solid state at -10°C to its liquid state at 1°C

= 10 x s₁{tm-(-10)} + 10L+ 10 x s₂ x (1 – t_m)

= 8(t_m+10) + 10L +10(1 – t_m) = 880

According to the problem,

8(t_m+ 10) + 10L+ 10(1 – t_m) = 880

or, 10L = 880 – 80- 10- 8t_m+ 10t_m = 790 +2t_m

or, L=79 + 0.2t_m (Proved).

Example 6. A piece of copper of mass 7.5 g at 27°C Is dropped into boiling liquid oxygen (boiling point = -183°C). Oxygen vapour thus formed occupies 1.83 L space at 20°C and at 750 minHg pressure. Calculate the latent heat of the vaporisation of oxygen. Specific heat of copper = 0.08 cal · g^-1 · °C^-1 and density of oxygen at STP 1.429 g · L^-1.
Solution:

A piece of copper of mass 7.5 g at 27°C Is dropped into boiling liquid oxygen (boiling point = -183°C). Oxygen vapour thus formed occupies 1.83 L space at 20°C and at 750 minHg pressure.

Let at STT, V₂ = volume of the oxygen vapour formed.

Given p₁ = 750 mmHg, T₁ = 273 + 20 = 293 K,

⇒ \(V_1=1.83 \mathrm{~L}, p_2=760 \mathrm{mmHg}, T_2=273 \mathrm{~K}\)

Now, \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\)

or, \(V_2=\frac{p_1 V_1 T_2}{T_1 p_2}\)

or, \(V_2=\frac{750 \times 1.83 \times 273}{293 \times 760}=1.68 \mathrm{~L}\)

∴ Mass of oxygen vapour formed = 1.68 x 1.429 g

Let L be the latent heat of the vaporisation of liquid oxygen. Heat taken for vaporisation = 1.68 x 1.429 x Leal

Heat given up by the piece of copper = 7.5 x 0.08(27 + 183) = 7.5 x 0.08 x 210 cal

∴ Heat supplied = heat absorbed

∴ 7.5 x 0.08 x 210 = 1.68 x 1.429 x L

or, L = \(\frac{7.5 \times 0.08 \times 210}{1.68 \times 1.429}=52.5 \mathrm{cal} \cdot \mathrm{g}^{-1}\)

Example 7. 100 g of water is raised from 24°C to 90°C with the help of steam. Determine the quantity of steam required for this purpose. Latent heat of steam = 540 cal · g^-1.
Solution:

100 g of water is raised from 24°C to 90°C with the help of steam.

Let the required quantity of steam = m g.

Amount of latent heat released by m g steam at 100°C = m x 540 = 540 m cal

Now, heat is released by this 100 g of condensed water at 100°C to reach 90°C = m x 1 x (100 – 90) = 10 m cal.

∴ Total heat released = 540 m + 10 m = 550 m cal

The amount of heat required by 100 g of water at 24 °C to rise up to 90°C = 100 x 1 x (90 – 24) = 6600 cal.

Heat lost = heat gained

∴ 550 m = 6600 or, m = 12

So, the quantity of steam required is 12 g

Example 8. A chunk of ice is continuously supplied with heat. After 2 s the ice begins to melt and in another 20 s the entire ice melts. Determine the initial temperature of ice. Specific heat of ice = 0.5 cal · g^-1 · °C^-1 and latent heat of fusion of ice = 80 cal · g^-1.
Solution:

A chunk of ice is continuously supplied with heat. After 2 s the ice begins to melt and in another 20 s the entire ice melts.

Let rate of supply of heat = x cal · s^-1, initial temperature of ice = -θ° C and amount of ice = mg

Since, the ice begins to melt after 2 s, we can assume that the temperature of ice becomes 0°C at that time.

So, m x 0.5 x [0 – (-θ)] = 2x……(1)

or, mdθ = 4x

In the next 20 s the ice melts completely

∴ m x 80 = 20x …..(2)

So, from (1) and (2) we get, 6 = 16

∴ The initial temperature of the ice was -16°C.

Example 9. What will be the result of extraction of 69000 cal of heat from 100 g of steam at 100°C? Latent heat of condensation of steam = 540 cal • g^-1
Solution:

The amount of heat extracted from 100 g of steam at 100°C to condense it to water at 100°C = 100 x 540 = 54000 cal.

The amount of heat extracted to bring down the temperature of this water from 100°C to 0°C = 100 x 1 x 100 = 10000 cal.

Total heat extracted in the two steps = 54000 + 10000 = 64000 cal

The additional amount of heat extracted = 69000 – 64000 = 5000 cal

This will freeze some amount of water at 0°C.

The amount of frozen ice = \(\frac{5000}{80}\) = 62.5 g.

Hence, finally, there will be a mixture of 62.5 g of ice and (100-62.5) = 37.5 g of water at 0°C.

Example 10. kg of Ice at -20°C is mixed with 5 kg of water at 20°C in a container of negligible thermal capacity. Calculate the amount of water finally left in the container. Specific heat of water = 1 kcal · kg^-1 °C^-1, specific heat of ice = 0.5 kcal · kg °C^-1 and latent heat of fusion of ice = 80 kcal · kg^-1.
Solution:

kg of Ice at -20°C is mixed with 5 kg of water at 20°C in a container of negligible thermal capacity.

Amount of heat given up by 5 kg of water to come at 0°C from 20°C = 5 x 1 x (20 – 0) = 100 kcal

Amount of heat taken by 2 kg ice at -20 °C to reach 0°C = 2 x 0.5 x {0 – (-20)} = 20 kcal

Amount of heat taken by 2 kg of ice at 0°C to melt into water at 0°C = 2 x 80 = 160 kcal

∴ Total amount of heat taken by 2 kg ice at -20°C to turn into the water at 0°C = 160 + 20 = 180 kcal

Obviously, the ice does not melt totally as the heat required to melt the total ice is greater than the heat given up.

∴ Amount of melted ice = \(\frac{100-20}{80}=\frac{80}{80}=1 \mathrm{~kg}\)

Hence, finally the amount of water in the container = 1 + 5 = 6 kg.

Example 11. A closed vessel is partly filled with water. While extracting air from the vessel by a pump, due to quick evaporation water freezes Into Ice. What fraction of the Initial mass of water can be frozen by this process? I .a tent heat of solidification of Ice = 00 cal · g^-1 and latent heat of vaporisation of water = 540 cal · g^-1.
Solution:

A closed vessel is partly filled with water. While extracting air from the vessel by a pump, due to quick evaporation water freezes Into Ice.

Let m₁ = mass of water frozen and m₂ = mass of water vaporised. Total mass of water (m) = m₁ + m₂.

Heat given out by m₁g of water for changing to Ice = 80 m₁ cal

And heat absorbed by m₂g of water for evaporation = 540m₂ cal

By calorimetric principle, 540m₂ = 80m₁

∴ \(m=m_1+\frac{80}{540} m_1=\frac{31}{27} m_1\)

∴ \(m_1=\frac{27}{31} m\)

∴ 27/31 part of the initial amount of water changes to ice.

Properties Of Bulk Matter – Change Of State Of Matter Synopsis

Latent heat: The amount of heat that should be applied to or extracted from a substance of unit mass for its change of state at a constant temperature, is called the latent heat for that particular change of state.

Melting point: The temperature at which a solid substance begins to melt under a definite pressure is called the melting point of that substance at that pressure.

Freezing point: The temperature at which a liquid substance begins to freeze under a definite pressure is called the freezing point of that substance at that pressure.

Eutectic temperature: The temperature at which a solution as a whole turns into a solid, is called the eutectic temperature of that solution.

• Latent heat of fusion of ice is 80 cal • g^-1.
• For substances which undergo contraction in volume on melting, the melting point decreases due to increase in pressure, i.e., they melt at a lower temperature.
• For substances which undergo expansion in volume on melting, the melting point increases due to increase in pressure, i.e. they melt at a higher temperature.
• The phenomenon of melting of ice under pressure and its resolidification when the pressure is released, is called regelation.

Evaporation: When vaporisation of a liquid occurs slowly from the exposed surface of the liquid at any temperature, the process is called evaporation.

• At a given temperature, if a closed space contains the maximum possible amount of vapour then that vapour is called saturated vapour. The pressure exerted by that vapour is called saturated vapour pressure.
• At a given temperature, if a closed space contains less amount of vapour than the maximum possible amount of vapour that it can contain, then the vapour is called unsaturated vapour. The pressure exerted by that vapour is called unsaturated vapour pressure.
• At constant temperature unsaturated vapour obevs Boyle’s law. It is seen that unsaturated vapour pressure is directly proportional to the temperature when volume remains constant, similar to the pressure law for gases.

Saturated vapour does not obey Boyles law nor the pressure law. The definite temperature for a gaseous substance, above which it cannot be liquefied by the application of pressure only, is called the critical temperature for that gaseous substance.

Boiling: When vaporisation of a liquid occurs rapidly throughout the whole of the liquid at a fixed temperature, the process is called boiling.

Sublimation: It is another form of vaporisation. When a solid turns into vapour directly without going through the liquid state, the process is known as sublimation.

Boiling point: The definite temperature at which a liquid boils under a fixed pressure is called the boiling point of that liquid.

Latent heat of vaporisation of water is 537 cal • g~1 .

The pressure and temperature at which the solid, liquid and gaseous states of a substance coexist in equilibrium is known as the triple point of the substance.

Properties Of Bulk Matter – Change Of State Of Matter Useful Relations For Solving Numerical Problems

During change of state, heat gained or lost by a substance, H = mL

[where, m =inass of the substance and L = latent heat for the change of state]

During change in temperature of a substance, heat gained or lost, H = mst

[where, s = specific heat of the substance, t = change in temperature]

Properties Of Bulk Matter – Change Of State Of Matter Very Short Answer Type Questions

Question 1. What is the heat gained or lost by a substance during its change of state at a constant temperature called?
Answer: Latent heat

Question 2. What is the unit of latent heat in CGS system?
Answer: cal • g^-1

Question 3. ‘Boiling point of water is higher at Darjeeling than that at Kolkata’ — state whether the statement is true or false.
Answer: False

Question 4. What is the latent heat of melting of ice in cal • g^-1?
Answer: 80

Question 5. What is the latent heat of the vaporisation of water in cal · g^-1? Answer: 540

Question 6. Due to increase in pressure on ice, its melting point _________
Answer: Decreases

Question 7. Name the phenomenon of melting of ice under pressure and its resolidification on withdrawal of pressure.
Answer: Regelation

Question 8. Name the process of vaporisation that occurs slowly from the upper surface of a liquid at any temperature.
Answer: Evaporation

Question 9. The boiling point of a solution is _______ than the boiling point of the solvent.
Answer: Higher

Question 10. Cooking becomes rapid in a pressure cooker, because the _________ of water increases due to increase of superincumbent pressure.
Answer: Boiling point

Properties Of Bulk Matter – Change Of State Of Matter Assertion Reason type

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Question 1.

Statement 1: In the pressure-temperature (p-T) phase diagram of water the slope of the melting curve is found to be negative.

Statement 2: Ice contracts on melting to water.

Answer: 1. Statement 1 is true, statement 2 is true statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: Melting of solid causes no change in internal energy.

Statement 2: Latent heat is the heat required to melt a unit mass of solid.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 3.

Statement 1: Water kept in an open vessel will quickly evaporate on the surface of the moon.

Statement 2: The day temperature at the surface of the moon is higher than boiling point of water.

Answer: 2. Statement 1 is true, statement 2 is true statement 2 is not a correct explanation for statement 1.

Question 4.

Statement 1: If 1 g of ice is mixed with 1 g of water at 80 °C, then the final temperature of the mixture is 0°C.

Statement 2: Ice melts.

Answer: 2. Statement 1 is true, statement 2 is true statement 2 is not a correct explanation for statement 1.

Question 5.

Statement 1: The specific heat of ice at 0°C is infinite.

Statement 2: When heat is supplied to ice at 0°C, the change in temperature is zero as the ice melts.

Answer: 1. Statement 1 is true, statement 2 is true statement 2 is a correct explanation for statement 1.

Properties Of Bulk Matter – Change Of State Of Matter Match Column 1 And Column 2.

Question 1. In a container of negligible mass ra g of steam at 100°C is added to 100 g of water that has a temperature 20 °C. If no heat is lost to the surroundings at equilibrium, match the items given in Column 1 with that in Column 2.

Answer: 1. C, 2. A, 3. D, 4. A

Question 2. Match the Column 1 with Column 2 according to data collected from the given graph.

Answer: 1.D, 2. A, 3. B, 4. C

Question 3. Match the Column 1 with Column 2 according to data collected from the given graph.

Answer: 1. C, 2. A, 3. D, 4. B

 

Unit 7 Properties Of Bulk Matter Chapter 8 Change Of State Of Matter Multiple Choice Questions And Answers

Question 1. The approximate value of latent heat of melting of ice in SI is

1. 0.336 J · kg^-1
2. 3.36 x 10⁵ J · kg^-1
3. 80 kcal · kg^-1
4. 80000 J · kg^-1

Answer: 2. 3.36 x 105 J · kg^-1

Question 2. The function of the latent heat during vaporisation is

1. Breaking the crystalline structures of the molecules of the substance
2. Restoring the crystalline structure of the molecules of the substance
3. Increasing the intermolecular separation of the substance
4. Decreasing the intermolecular separation of the substance

Answer: 3. Increasing the intermolecular separation of the substance

Question 3. Different articles of iron can be prepared by casting of molten cast iron because

1. The latent heat of solidification of iron is very high
2. The latent heat of solidification of iron is very low
3. Cast iron decreases in volume during solidification
4. Cast iron increases in volume during solidification

Answer: 4. Cast iron increases in volume during solidification

Question 4. The ratio of the heat required to increase the temperature of some water from 0°C to 50°C to the heat required to convert the ice of same mass into steam is

1. 5/6
2. 1/8
3. 16/31
4. 5/72

Answer: 4. 5/72

Question 5. 300 g of water at 80°C is poured on a big piece of ice at 0°C. The mass of the molten ice will be

1. 80 g
2. 30 g
3. 800 g
4. 300 g

Answer: 4. 300 g

Question 6. What will be the final temperature if 90 g of water at 20°C is mixed with 10 g of ice at 0°C?

1. 20°C
2. 2°C
3. 10°C
4. 18°C

Answer: 3. 18°C

Question 7. The final temperature of the mixture when 1 g of water at 100°C is mixed with 1 g ice at 0°C will be

1. 0°C
2. 10°C
3. 90°C
4. 100°C

Answer: 2. 10°C

Question 8. The final temperature of the mixture when lg of steam at 100°C be mixed with lg of ice at 0°C will be

1. 0°C
2. 10°C
3. 90°C
4. 100°C

Answer: 4. 100°C

Question 9. Freezing point of water is 0°C. If some salt is dissolved in water then freezing point of the mixture will be

1. 0°C
2. More than 0°C
3. Less than 0°C
4. 10°C

Answer: 3. 10°C

Question 10. More time is required to cook food at Darjeeling, because

1. Atmospheric temperature is less there
2. Humidity of atmospheric is less there
3. Atmospheric pressure is less there
4. Atmospheric pressure is more there

Answer: 3. Atmospheric pressure is more there

Question 11. The relation between melting point and freezing point of a noncrystalline substance is

1. Both the temperature are equal
2. Melting point > freezing point
3. Melting point < freezing point
4. Melting point may be higher or lower than its freezing point depending on the nature of the substance

Answer: 2. Melting point > freezing point

Question 12. To construct an electrical fuse,

1. An alloy is used because its melting point is higher
2. An alloy is used because its melting point is lower
3. An alloy is never used because its melting point is lower
4. An alloy is never used as its melting point is higher

Answer: 2. An alloy is used because its melting point is lower

Question 13. Melting point of ice, due to increase in pressure on it,

1. Decreases, because excess pressure helps melting
2. Increases, because excess pressure helps melting
3. Decreases, because excess pressure opposes melting
4. Increases, because excess pressure opposes melting

Answer: 1. Decreases, because excess pressure helps melting

Question 14. The rate of melting of the lower surface of an iceberg is more than its upper surface because

1. The surrounding temperature of the lower surface is higher
2. The amount of impurities at the lower surface is greater
3. More pressure acts on the lower surface so that melting point of ice decreases there
4. The structure of ice at the upper surface is different

Answer: 3. More pressure acts on the lower surface so that melting point of ice decreases there

Question 15. The ice line is the curve which shows the variation of melting point of ice with pressure. This graph

1. Has a positive gradient
2. Has a negative gradient
3. Is parallel to the pressure axis
4. Is parallel to the temperature axis

Answer: 2. Has a negative gradient

In this type of questions more than one options are correct

Question 16. With increase in pressure on a substance

1. Boiling point may be elevated
2. Boiling point may be depressed
3. Melting point may be elevated
4. Melting point may be depressed

Answer:

1. Boiling point may be elevated

3. Melting point may be elevated

4. Melting point may be depressed

Question 17. We can get vapour state of a substance through the process of

1. Boiling
2. Evaporation
3. Sublimation
4. Regelation

Answer:

1. Boiling
2. Evaporation
3. Sublimation

Question 18. When m g of water at 10°C is mixed with m g of ice at 0°C, which of the following statements are false?

1. The temperature of the system will be given by the equation
   m x 80 + m x 1 x (T-0) = m x 1 x (10 – T)
2. Whole of ice will melt and temperature will be more than 0°C but lesser than 10°C
3. Whole of ice will melt and the temperature will be 0°C
4. Whole of ice will not melt and temperature will be 0°C

Answer:

1. The temperature of the system will be given by the equation
   m x 80 + m x 1 x (T-0) = m x 1 x (10 – T)
2. Whole of ice will melt and temperature will be more than 0°C but lesser than 10°C
3. Whole of ice will melt and the temperature will be 0°C

Question 19. Which of the following statements are true?

1. Water in a test tube can be made to boil by placing it in a bath of boiling water
2. Heat cannot be stored in a body
3. With increase in pressure melting point decreases
4. Vapour can be directly converted into solid

Answer:

2. Heat cannot be stored in a body

4. Vapour can be directly converted into solid

Properties Of Bulk Matter – Change Of State Of Matter Vaporisation Condensation

The gaseous state ot a liquid is called the vapour of that liquid. The process by which a liquid changes into vapour is called vaporisation. The opposite process is called condensation, i.e. the conversion of vapour into liquid is
called condensation or liquefaction.

Just as a solid substance absorbs latent heat when it converts into a liquid, a liquid also absorbs latent heat when it converts into vapour.

The amount of heat absorbed by unit mass of a liquid to convert into vapour is called latent heat of vaporisation. This value depends on the temperature at which the vaporisation occurs.

Read and Learn More: Class 11 Physics Notes

Vaporisation of liquids may take place in two ways:

1. Evaporation and
2. Boiling.

 

1. The gradual change of a liquid to its vapour state, at any temperature, from the surface of the liquid is called evaporation.
2. The rapid change of a liquid to its vapour state, at a certain temperature, from the whole volume of the liquid, is called boiling. This particular temperature depends on the pressure on the liquid and remains constant until the whole liquid transforms into vapour.

Sublimation: It is another process of vaporisation where a solid changes directly to its vapour state, without passing through the liquid state. The vapour when condensed, regains the solid state directly. Camphor, iodine, napthalene etc. change into vapour under normal temperature.

Evaporation: If we store water in a wide-brimmed pot, it is noticed that the level of water decreases in a couple of days. This happens due to evaporation from the surface of the water. During summers, small ponds and canals dry up because of evaporation.

• This process is slow but continuous and can take place at all temperatures. The rate of evaporation increases with increase in temperature. Wet clothes dry up by the process of evaporation of water.
• All liquids evaporate somewhat under ordinary temperatures. Liquids, which evaporate at ordinary temperature at a very high rate, are classified as volatile liquids. Ether, alcohol, chloroform, carbon tetrachloride are examples of volatile liquids.
• Rate of evaporation for some liquids, at ordinary temperature, is very low. Such liquids are termed as nonvolatile liquids. Mercury, glycerine etc. belong to this group.

Factors affecting the rate of evaporation:

1. Nature of the liquid: Different liquids have different rates of evaporation. The liquids with low boiling points, also called volatile liquids, have high rates of evaporation. So, spirit, ether, petrol, etc. evaporate quickly.

2. Surface area of liquid: Rate of evaporation increases with the increase in surface area of the liquid. Due to larger surface area, a higher number of molecules of the liquid can leave the surface at a time. The surface area of wet clothes are increased by spreading them as much as possible; this increases the rate of evaporation and the clothes dry up faster.

3. Dryness of air: The drier the air (i.e., the lower the humidity), the faster the evaporation. The air is less humid in winter than during monsoon. So wet clothes dry faster in winter than during monsoon.

4. Atmospheric pressure: Higher the pressure on the liquid surface, lower the rate of evaporation. The converse is also true. For this reason, evaporation is very rapid in vacuum.

5. Temperature of liquid and the surrounding air: Higher the temperature of the liquid and the surrounding air, higher the rate of evaporation. Hence, ponds, tanks etc. dry up quickly in summer.

6. Flow of air: Flow of air over the liquid surface increases the rate of evaporation. Continuous flow of air shifts moist air from over the surface of the liquid replacing it with dry air that helps evaporation. So, perspiration dries up faster under a fan.

Cooling due to evaporation: A liquid needs some latent heat to change into its vapour state. If heat is not supplied from outside, liquid absorbs the latent heat from its own body and from the surroundings during evaporation. Hence, the liquid and the surroundings get cooled.

Cooling due to evaporation Example:

1. Ether or spirit, dropped on our skin, produces a cooling effect. Such a volatile liquid, evaporates fast by absorbing latent heat from our body and this producing this cooling effect.
2. Earthen pitcher keeps water cooler than a metal pitcher. An earthen pitcher has a large number of pores on its surface. Water comes out through the pores and evaporates, collecting the necessary latent heat from the pitcher and also from the water in the pitcher.
   • This causes a fall in the temperature of the water in the pitcher. Pores of old earthen pitchers get blocked by dust particles and so water kept in them do not cool down as much.
   • Water in metal or glass vessels do not cool down much because of the lack of pores on the surface of the container. Evaporation can only take place from the top of the liquid surface. So the water inside does not cool down as much.
3. To keep a room cool in summer, vetiver grass (khas khas or khus khus) mattings with water sprayed on the surface, are hung on doors and windows. The sprayed water evaporates and collects the latent heat from the air inside, thereby cooling the room.
4. Cloth pieces soaked in water are spread on the forehead of a person with high fever. Water in the wet cloth piece evaporates, taking the necessary latent heat from the forehead. So it helps in lowering the body temperature and hence the fever.
5. Wet clothes, when allowed to dry on the body, often cause cough and cold. Water of the wet clothes evapo¬rates taking its latent heat from the body, thus lowering the body temperature; this is the reason for catching cold.
6. If we stand under a fan (or in a breeze) after being covered in perspiration we immediately experience a cooling sensation. This is because perspiration evaporates rapidly under a fan and as a result, it takes away latent heat from the skin as well as the body.
7. Water is poured on roads, room floors or rooftops to lower the temperature during summer. This water, while evaporating, absorbs latent heat from the roads, floors or rooftops, thereby creating a cooling effect.

• Saturated Vapour and Saturated Vapour Pressure
• Unsaturated Vapour and Unsaturated Vapour Pressure

Vaporisation takes place at any temperature from the upper surface of a liquid. In a closed container when the liquid changes to vapour, the container gradually becomes filled with that vapour.

• The vapour thus formed, exerts pressure on the surface of the container like any other gas. This pressure is termed as vapour pressure.
• As the vapour formed due to evaporation in the closed container increases, vapour pressure also increases. There is a limit up to which a closed space of fixed volume can hold a certain amount of vapour at a fixed temperature. When these closed container holds maximum amount of vapour at a certain temperature that it can hold, the container is saturated with vapour.
• In this condition, the vapour container over the liquid surface in the container is called I saturated vapour and the pressure exerted by that vapour on the container is called saturated vapour pressure.
• On the other hand, if the vapour present is less than the amount that a closed space can hold maximum at that temperature, the space is called unsaturated with that vapour.
• In this state, the amount of vapour that is present in the container is called unsaturated vapour, and the pressure it exerts on the surface of the container is known as unsaturated vapour pressure.

Saturated vapour and saturated vapour pressure: The maximum possible amount of vapour that a closed container can hold at a specific temperature is called saturated vapour and the pressure exerted by this vapour is known as saturated vapour pressure.

Unsaturated vapour and unsaturated vapour pressure: If the amount of vapour present in a closed container is less than the maximum possible amount that the closed container can hold at a specific temperature, then the vapour is known as unsaturated vapour and the pressure it exerts is called unsaturated vapour pressure.

It is important to note that though it is possible to get a fixed value of saturated vapour pressure at any certain temperature, unsaturated vapour pressure is devoid of any fixed value. Also, the saturated vapour pressure, of different liquids at a fixed temperature are different.

Change in Volume of Vapour at Constant Temperature:

Unsaturated vapour: Unsaturated vapour follows Boyle’s law at a constant temperature. Hence, in a closed space

1. Unsaturated vapour pressure decreases with the increase in space (i.e., Volume)
2. When the volume of the space is decreased, initially the vapour pressure increases in inverse proportion. As the reduction in space continues, the capacity of holding vapour in the space also decreases and unsaturated vapour finally changes to saturated vapour. Now, even on further reduction of volume in space, the vapour pressure does not follow boyle’s law any more.

Saturated vapour: Saturated vapour pressure at a fixed temperature is independent of the volume of vapour i.e. saturated vapour does not obey Boyle’s law.

• The observations can be represented graphically as shown. Curved portion of the graph, DC shows that unsaturated vapour obeys Boyle’s law. At point C, the vapour becomes saturated.
• Its pressure remains constant even if its volume i$ decreased more and the saturated vapour gradually condenses into liquid under pressure. CB part of the graph represents this change.
• Obviously, CB is parallel to the volume axis. At point B, the entire vapour condenses into liquid. No appreciable change in the volume of the liquid is noticed with the increase in pressure.
• This is because liquid is almost incompressible. Hence, the BA pan of the graph is almost parallel to the pressure axis.

Change in Temperature of Vapour at Constant

Unsaturated vapour: Unsaturated vapour pressure is directly proportional to the temperature when volume remains constant, similar to the pressure law for gases.

Saturated vapour: Saturated vapour pressure also increases with the increase in temperature at a constant volume, but not as per pressure law. The nature of variation between saturated vapour pressure of water and temperature can be represented graphically as shown.

Initially, vapour pressure increases at a slower rate. Soon the rate of increase in pressure becomes very high. Two important points are to be noted here.

1. Saturated vapour pressure of water at 0°c is not zero, but about 0.4 cm of mercury pressure and
2. At 100°c, the saturated vapour pressure of water is 76 cm of mercury which is normal atmospheric pressure. For other pure liquids too, the graph is almost the same.

Differences between Saturated and Unsaturated Vapours

1. At a constant temperature, if the amount of any vapour and amount of the liquid in contact with that vapour do not change with time then it can be said that the vapour and the liquid are in equilibrium to each other.

2. Saturated vapour pressure of a liquid is equal to its superincumbent pressure during boiling. For example at 98°C the saturated vapour pressure of water is 707.3 mm, i.e., at the atmospheric pressure of 707.3 mm of mercury, water will boil at 98°C.

Saturated vapour pressure (SVP) of water at different temperatures (Regnaulfs list)

• Critical Temperature
• Gas and Vapour

Critical temperature: It has been observed that when a gas is cooled below a certain temperature, it can be converted to its liquid state by application of pressure alone. When the temperature of the gas is above that temperature, the gas cannot be compressed to liquid.

This fixed temperature for a gas is called the critical temperature. The pressure applied to change a gas to its liquid state at critical temperature is called its critical pressure. Every gas has its characteristic critical values of temperature and pressure.

Gas and vapour: A gas below its critical temperature is a vapour and vapours can be compressed to the liquid state. But when the temperature of a gaseous matter is above its critical temperature, it is a gas. Naturally, a gas cannot be liquefied only by application of pressure.

• Critical temperatures of oxygen and hydrogen are -119°C and -24 °C respectively. Hence, at room temperature, these are permanent gases and cannot be liquefied by applying pressure alone.
• But, the critical temperatures for carbon dioxide, ammonia, sulphur dioxide are 31°C, 132.2°C and 157.2°C respectively, and are above the normal room temperature.
• So, they can be compressed to liquid at room temperature and are therefore called vapours. It is to be noted that to liquefy a gas it is, at first, cooled below its critical temperature and then appropriate amount of pressure is applied to convert it to liquid.

Boiling: The temperature of a liquid increases on heating and vapour starts rising from the surface. When the liquid reaches a particular temperature, it gets vigorously agitated and the whole volume of the liquid starts to transform into vapour.

This state of the liquid is called boiling. This temperature remains constant until the entire liquid changes into vapour, and is called the boiling point of the liquid. Heat supplied during boiling is entirely used for transition of liquid into its vapour and here is no rise in the temperature.

Boiling Definition: The temperature at which the whole volume of a liquid starts transforming into its vapour state rapidly, under a fixed pressure is called the boiling point of that liquid at that pressure.

Different liquids have different boiling points. The boiling point of any liquid depends on the pressure on the liquid surface. At standard atmospheric pressure, the temperature at which a liquid boils is called its normal boiling point. Every liquid has a normal boiling point. For example, the normal boiling point of water is 100°C, i.e., at standard atmospheric pressure, water boils and transforms into its vapour at 100°C.

Normal boiling points of a few liquids

Differences between Evaporation and Boiling

Latent Heat of Vaporisation and Latent Heat of Condensation: We know that the heat applied during boiling does not increase the temperature of the liquid. This heat transforms liquid into its vapour. The heat required to transform unit mass of a liquid into its vapour is called latent heat of vaporisation.

Latent heat of vaporisation: The amount of heat required by unit mass of a liquid to change into its vapour at a constant temperature, is the latent heat of vaporisa¬tion of the liquid.

Similarly, if we cool a gas, it will start condensing on reaching the boiling point. Until the whole gas condenses, the temperature remains constant. Clearly, a gas loses heat during condensation. The amount of heat lost by unit mass of a gas during condensation is called latent heat of condensation.

Latent heat of condensation: The amount of heat extracted from unit mass of a vapour to change it into its liquid state at a constant temperature, is the latent heat of condensation of the vapour.

• For any substance, the latent heat of vaporisation is equal to the latent heat of condensation and is denoted by L. Thus if a quantity H of heat is given or extracted to change the state of mass m, for vaporisation or condensation respectively then H = mL. Unit of L in CGS system is cal · g^-1, and in SI it is J · kg^-1
• The latent heat of vaporisation of water or latent heat of steam is 537 cal · g^-1; it means that 537 cal of heat is required by 1 g of water at 100°C to change into 1 g of steam at 100°C. Again, 537 cal heat is to be extracted from 1 g of steam at 100°C to change it into 1 g of water at 100°C.
• So, it is clear that, 1 g of steam at 100°C can transfer 537 cal more heat than 1 g of water at 100° C. Hence, burns caused by steam at 100°C are more severe than those caused by boiling water at 100°C.
• Besides, the latent heat of vaporisation of water is more than that of any other liquid. So steam at 100°C is a very good warming agent, in cold countries steam is used, instead of hot water to keep the houses warm.

Values of latent heat of vaporisation of water:

In CGS system, L = 537cal · g^-1

In SI, L = 537 x 4.2 x 10³ = 2.26 x 10⁶ J. kg^-1

Latent heat of vaporisation of a few liquids

Latent heat of evaporation and latent heat of boiling are same for a particular liquid at a particular temperature. This is why both these quantities are referred to as latent heat of vaporisation. But in general, evaporation and boiling do not take place at the same temperature.

In case of water, the latent heat of vaporisation increases with decrease in temperature. For example, at 100°C, during boiling of water latent heat is 537cal-g^-1. But when water is evaporated from ponds, rivers and other water bodies at 40°C or below, the latent heat of vaporisation is higher than 537 cal • g^-1.

Properties Of Bulk Matter – Change Of State Of Matter Vaporisation Condensation Numerical Examples

Example 1. How much heat is required to convert 1 g of ice at -10°C to steam at 100°C? Specific heat of ice = 0. 5 cal · g^-1 · °C^-1; latent heat of fusion of ice = 80 cal · g^-1; latent heat of vaporisation of water = 540 cal · g^-1
Solution:

Specific heat of ice = 0. 5 cal · g^-1 · °C^-1; latent heat of fusion of ice = 80 cal · g^-1; latent heat of vaporisation of water = 540 cal · g^-1

Heat required to bring 1 g of ice from -10°C to 0°C

= 1 x 0.5 x {0 – (-10)} = 5 cal

Heat required to convert 1 g ice at 0°C to water at 0°C = 1 x 80 = 80 cal

Heat required to bring that water to 100°C 1 x 1 x (100-0) = 100 cal

To convert water at 100°C to steam at 100°C, heat required = 1 x 540 = 540 cal

∴ The total heat required to convert the ice to steam = 5 + 80 + 100 + 540 = 725 cal.

Example 2. What will be the consequence of extracting 64800 cai of heat from 100 g of steam at 100°C? Latent heat of condensation of steam = 540 cal · g^-1; latent heat of fusion of ice = 80 cal · g^-1.
Solution:

Latent heat of condensation of steam = 540 cal · g^-1; latent heat of fusion of ice = 80 cal · g^-1.

The heat extracted from 100 g of steam at 100°C to form water at 100°C = 100 x 540 = 54000 cal.

Then, heat extracted to bring that water to 0°C = 100 x 100 = 10000 cal

The total heat extracted in these two stages = 54000 + 10000 = 64000 cal

The remaining amount of extracted heat = 64800 – 64000 = 800 cal. It freezes some amount of water into ice.

∴ The amount of ice formed = 800/80 = 10 g.

∴ On extracting 64800 cal of heat, the specimen will consist of 10 g of ice at 0°C and (100 – 10) g i.e., 90 g of water at 0°C.

Example 3. 100 g of steam is passed through a mixture of 1 kg ice and 1 kg water. The entire amount of steam is converted into water. What is the final temperature of the mixture? What amount of ice will melt? The latent heat of fusion of ice is 80 cal • g^-1 and that of condensation of steam is 540 cal • g^-1.
Solution:

100 g of steam is passed through a mixture of 1 kg ice and 1 kg water. The entire amount of steam is converted into water.

The initial temperature of the mixture of ice and water = 0°C

The heat released by 100 g steam at 100°C to be converted to water at 100°C = 100×540 = 54000 cal

Heat released by that water from 100°C to 0°C = 100 x 100 = 10000 cal

∴ Total heat released by steam = 54000 + 10000 = 64000 cal

The heat required to melt 1 kg ice = 1000 x 80 = 80000 cal

So, all the ice will not melt as the total heat released is less.

So, the amount of ice that melts = 64000/80 = 800 g

and the final temperature of the mixture will be 0°C.

Example 4. Divide 1 kg of water at 5°C in two parts such that the heat released in freezing one part into ice at 0 C can be used to convert the other part into steam. Latent heat of the solidification of water and that of the vaporisation of water are 80 cal · g^-1 and 540 cal · g^-1 respectively.
Solution:

Latent heat of the solidification of water and that of the vaporisation of water are 80 cal · g^-1 and 540 cal · g^-1 respectively

Solution: Let the mass of the first part of water = x g.

∴ Mass of the remaining part of water = (1000 – x) g

The heat released by xg water for converting into ice = x x (5 – 0) + x x 80 = 85x cal

The heat required to convert (1000 – x) g of water to steam = (1000 -x)x (100 – 5) + (1000 – x) x 540 = (1000-x) x 635 cal

∴ 85x = (1000-x) x 635

or, 720x = 635000

or, \(=\frac{63500}{72}=881.9 \mathrm{~g}\)

∴ Mass of the first part = 881.9 g and that of the other part = 1000-881.9 = 118.1 g.

Example 5. A certain amount of water is heated from CTC to 100°C in an electric kettle. It takes 15 min to raise the temperature and 80 min to completely convert the.water to steam. What is the latent heat of the vaporisation of water?
Solution:

A certain amount of water is heated from CTC to 100°C in an electric kettle. It takes 15 min to raise the temperature and 80 min to completely convert the.water to steam.

Let amount of heat generated in the kettle every minute = H cal.

Amount of water contained in the kettle = x g

∴ According to the given data,

15H = \(x(100-0) or, \text { or, } \frac{H}{x}=\frac{100}{15}=\frac{20}{3} \mathrm{cal} \cdot \mathrm{g}^{-1}\)

Again, in case of vaporisation, 80 H – x x L [where L = latent heat of steam)

or, 80\(\frac{H}{x}\)= L

or, L = 80 \(\times \frac{20}{3}=533.3 \mathrm{cal} \cdot \mathrm{g}^{-1}\)

∴ Latent heat of steam = 533.3 cal · g^-1.

Factors Influencing Boiling Point

1. Nature of the liquid: The boiling point of a liquid depends on the nature of the liquid. Different liquids have different boiling points. More volatile a liquid, the less its boiling point is.

2. Pressure on the liquid: The boiling point of a liquid depends on the pressure on the liquid. If the pressure decreases, boiling point decreases and vice versa.

3. Presence of impurities in the liquid: Boiling point of a pure liquid increases in presence of dissolved impurities. Saline water has a boiling point about 9 °C higher than pure water at the same pressure. Suppose, there is a possibility’ of presence of dissolved impurities in a liquid.

In this case, to determine the boiling point of the pure liquid, a thermometer is held in the vapour right above the liquid (instead of immersing it in the liquid). This works because if pressure on the liquid remains unchanged, the temperature of the vapour and the boiling point of the pure liquid (at the same pressure) are the same.

Effect of Pressure on Boiling Point: Boiling point of a liquid depends on the pressure on it. When pressure is reduced, the boiling point decreases. If the pressure on a liquid surface is increased, the boiling point of the liquid increases.

During vaporisation of any liquid, its volume increases. The pressure on the liquid opposes this expansion in volume. So to boil the liquid, it needs to be brought to a higher temperature.

Effect of Pressure on Boiling Point Application:

1. Decrease in the boiling point with the decrease in pressure—this property of a liquid is utilised in different cases. For example,

1. In preparing a concentrated solution of hydrogen peroxide,
2. Making of condensed milk,
3. Forming sugar crystals from a sugar solution etc.

2. Increase in the boiling point with the increase in pres¬sure—this property is also utilised in different cases. For example,

1. Cooking in a pressure cooker,
2. Making paper pulp from sawdust and caustic soda,
3. Making artificial silk,
4. Sterilization of surgical instruments, bandages etc.,
5. Preservation of food materials in tin containers,
6. Extraction of pure alumina from bauxite, etc.,
7. Modem electric power plants utilise water at high temperatures by exerting high pressure.

Effect of altitude on boiling point: The atmospheric pressure decreases with increase ip altitiicje from the earth’s surface. If the altitude from the earth’s stirface is not very high, the decrease in pressure is about 85 mm of mercury for every kilometer rise in altitude.

• We know that the boiling point of a liquid depends on the pressure on its surface. So, the boiling point decreases with the rise in altitude. As the atmospheric pressure is low on mountains, water boils at a temperature lower than 100°C.
• So it is harder to boil anything on mountains. It has been calculated that on the peak of Mt. Everest (about 9 km high) water boils at only 70°C. At Darjeeling (about 2.2 km high) the boiling point of water is 93.6°C.

Determination of altitude from boiling point: The altitude of a place or the difference in altitudes of two places can be determined by measuring the boiling point of water.

Let the atmospheric pressures at the base (A) and at the top (B) of a hill be p_A and p_B respectively. The boiling points of water at A and at B are determined using a pressure hypsometer. Let these be T_A and T_B respectively.

The normal boiling point of water is 100°C. Although the vapour pressure variation with temperature is non-linear in nature, the boiling point variation can be approximated near 100°C by an empirical result.

For a 2.7 cm Hg rise or fall of pressure, the normal boiling point of water rises or falls by 1°C approximately.

Let pressure difference between A and B be h cm Hg.

∴ h = p_A -p_B=2.7(T_A-T_B)…(1)

Now, Hρg = h x 13.6 x g

[where p is the average density of air]

H = \(\frac{h \times 13.6}{\rho}\)

We get from (1) and (2)

H = \(\frac{2.7\left(T_A-T_B\right) \times 13.6}{\rho} \mathrm{cm}\)

∴ H = \(\frac{2.7\left(T_A-T_B\right) \times 13.6}{\rho} \times 10^{-2} \mathrm{~m}\)

Laws of Boiling: The facts related to boiling can be expressed in the form of the following laws. These law’s are known as laws of boiling.

1. Every liquid has a natural boiling point. The tempera¬ture at which a liquid starts boiling at normal atmospheric pressure is its normal boiling point. During boiling, the temperature remains constant until all the liquid changes to vapour.
2. Boiling point increases with the increase in pressure on die liquid and decreases when the pressure is lowered.
3. Boiling point of a liquid increases due to the presence of dissolved substances. But the temperature of the vapour above the liquid remains equal to the boiling point of the pure liquid.

Comparison Between Melting And Boiling Similarity:

1. Both melting and boiling produce change of state at respective fixed temperatures, called melting point and boiling point, and these temperatures remain constant as long as the change continues.
2. In both processes heat is absorbed.
3. Both processes are affected by any change in pressure on the substance. In other words, both melting point and boiling point of a substance depend on pressure on the substance.
4. In both processes there occur changes in volume.
5. Unstirred pure liquid can be gradually cooled below its melting point and it can still retain its liquid state. This is called supercooling. Similarly, unstirred pure liquid can be gradually heated above its boiling point and it can still retain its liquid state. This is called superheating. However both the states are very unstable. Even the slightest change in environment can change these states.

Comparison Between Melting And Boiling Dissimilarity:

1. Freezing point or melting point of a solution is always less than that of the pure solvent, but boiling point of a solution is always more than that of the pure solvent.
2. The volume of a solid may increase or decrease on melting, depending on the material. However, the volume of all liquids increase on boiling.

Properties Of Bulk Matter Change Of State Of Matter Fusion And Solidification

Fusion or Melting: When heat is applied to a solid, initially its temperature rises. After reaching a particular temperature, the solid begins to melt. Until the whole substance converts to a liquid, the temperature does not change thought heat is continuously applied.

As soon as the solid completely melts, any further application of heat raises the temperature of the liquid again. The variation of temperature (due to application of heat at a constant rate) with time is shown in the graph.

Point A represents the initial temperature of the solid. On heating, it continues to be in solid state but its temperature increases. The line AB represents this stage.

Point B corresponds to a temperature θ where melting starts. The temperature remains at θ up to point C even though the heating continues. The part BC in the graph represents this stage. At this stage the specimen remains as a mixture of its solid and liquid forms θ is called the melting point of the solid.

Read and Learn More: Class 11 Physics Notes

Point C represents the stage where the whole mass of the solid has changed to liquid, and on further heating, a rise in temperature is noticed. This is represented by the part CD of the graph.

Solidification or freezing: On absorbing heat, a solid changes into a liquid and on losing heat, a liquid changes into a solid. Freezing, therefore, can be considered as the reverse process of melting.

When heat is extracted from a liquid, its temperatilre decreases initially. On reaching a particular temperature, the liquid begins to freeze. Until the whole liquid freezes, its temperature remains constant despite extraction of heat.

Once the whole mass of liquid solidifies, further extraction of heat results in cooling of the solid substance. A graphical representation of the change in temperature of the liquid (due to extraction of heat at a constant rate) with respect to time is shown.

Point A denotes the initial temperature of the liquid. The line BC represents solidification at constant temperature θ, which is called the freezing point of the liquid and CD represents the cooling of the solid thus formed.

• Melting Point
• Freezing Point

Melting point: At a fixed pressure, the temperature at which a solid starts melting on absorbing heat, is called its melting point at that pressure.

Freezing point: At a fixed pressure, the temperature at which a liquid starts freezing into a solid by losing heat, is called its freezing point at that pressure.

• Melting point or freezing point for different substances are different. Melting point or freezing point of a substance depends on the superincumbent pressure. Melting point or freezing point of a substance at standard atmospheric pressure is called its normal melting point or normal freezing point.
• Melting point and freezing point are exactly equal and fixed only for a pure crystalline substance. A noncrystalline substance like glass, wax, fat, butter, pitctar etc., does not have any fixed freezing or melting point.
• This type of substance has a temperature range within which it starts melting. (For example, Butter melts within 28 °C to 33°C whereas freezes within 20°C to 23°C).
• On heating, non-crystalline substances, first attain a viscous stage which is neither purely a liquid state nor purely a solid state. So these substances, do not have any fixed melting or freezing point.

In pure substances and mixtures of substances do not have fixed melting points. Again, a few substances like magnesium oxide (MgO), calcium oxide (CaO) etc. do not melt at any temperature.

Normal freezing/melting points of some substances

Melting point of an alloy: Melting point of an alloy is generally less than that of its individual constituents. Solder, an alloy of lead and tin, has a melting point of 180°C, lower than the melting points of tin (232°C) and lead (327°C). Wood’s metal (an alloy of tin, lead, bismuth and cadmium) has a melting point of 60.5°C, whereas its constituents have melting points greater than 60.5°C.

An alloy of lead, tin and bismuth, called Rose’s metal, melts at 94.5°C, while none of its constituents has a melting point lower than 210°C. Lower melting point of all alloy finds its application in the construction of fuse wires an electrical circuit. Such alloys are also used in fire alarms and circuit breakers.

Freezing point Of a solution: Any pure liquid has a fixed freezing point. When a solute is dissolved in it, the freezing point of the solution becomes lower than that of the pure liquid. For example, while the normal freezing point of pure water is 0°C, saline sea water freezes at -2°C.

If the amount of salt in saline water increases, freezing point of the solution decreases. It has been observed that the lowest freezing point of saline water is -22 °C when salt and water are in the mass ratio of 1: 3. So, water freezes in lakes, tanks etc. in cold countries, but sea water does not freeze easily. For the same reason, freezing point of milk is a little less than that of pure water.

• In cold countries to prevent bursting of the radiator tube of a car due to the freezing of water, glycerine or glycol is mixed with water. As a result, the freezing point decreases and water does not freeze even at 0°C. Glycerine, glycol etc. are called antifreeze substances.
• When a solution is gradually cooled and the solution starts solidifying, the pure solvent gets separated in crystalline form, from the solution. Hence, the proportion of solute gradually increases in the rest of the solution and the solution becomes denser.
• At the same time the freezing point of the solution also decreases. If the cooling process is continued, then more solvent, in crystalline form, gets separated from the solution. Finally, at a certain temperature, the rest of the solution condenses as a whole into a solid. This particular temperature is called the eutectic temperature.

Freezing point Of a solution Definition: The temperature at which a solution as a whole condenses into a solid, is called the eutectic temperature of that solution.

Following the method given above both salt and pure water can be prepared. As the temperature of saline water starts falling below 0°C, pure ice gets separated from the solution. It can be melted to obtain pure water.

In cold countries, salt is prepared from seawater. Sea water starts freezing at -2°C. At this temperature, some amount of pure water gets separated as ice. Hence, the amount of salt increases in the rest of sea water. This seawater is vaporised to obtain salt.

Freezing mixture: Freezing mixture is a mixture of salt and ice. For instance, the temperature of a mixture consisting of 3 parts crushed ice and 1 part common salt is -22CC, and that of a mixture of 2 parts ice and 3 parts decreased to calcium chloride is almost -50°C.

Practical uses of freezing mixture: For preservation and transport of food materials like fish, and meat, that decay easily due to putrefaction, freezing mixture is extensively used. It is also used for making ice cream, kulfi etc. and in laboratories to create low temperatures.

Latent Heat of Fusion Definition: The temperature at which a solution as a whole condenses into a solid, is called the eutectic temperature of that solution.

Similarly, the amount of heat that needs to be extracted to change a liquid of unit mass into its solid state, at a constant temperature, is called the latent heat of solidification of that liquid.

The latent heat of fusion and the latent heat of solidification of a substance are equal. Latent heat is generally denoted by L.

If H is the quantity of heat that is to be supplied to or extracted from a mass m of a substance to melt or solidify it then H = mL.

unit of heat unit of mass = \(\frac{\text { unit of heat }}{\text { unit of mass }}\)

Units of Latent Heat:

• CGS System: cal · g^-1
• SI: J · kg^-1

∴ \(1 \mathrm{cal} \cdot \mathrm{g}^{-1}=\frac{4.2 \mathrm{~J}}{10^{-3} \mathrm{~kg}}=4200 \mathrm{~J} \cdot \mathrm{kg}^{-1}\)

Latent heat of fusion of ice or latent heat of ice is 80 cal · g^-1: it means that to melt 1 g of ice at 0 °C into 1 g of water at 0 °C, 80 cal heat is to be supplied, or, to solidify 1 g of water 0°C into 1 g of ice at 0°C, 80 cal heat is to be extracted.

Latent heat of ice = 80 cal · g^-1 = 80 x 10³ cal · kg^-1

= 80 x 10³ x 4.2 J · kg^-1

= 3.36 x 10⁵ J · kg^-1

Therefore, in SI, the latent heat of ice is 3.36 x 10⁵ J · kg^-1.

Latent heat of fusion some substances

Properties Of Bulk Matter – Change Of State Of Matter Fusion And Solidification Numerical Examples

Example 1. A copper calorimeter of mass 300 g contains 270 g of water at 30° C. Now 20 g of ice at -10°C is added to it. What will be the final temperature? Specific heat capacity of copper = 0.1 cal · g^-1 · °C , specific heat capacity of ice = 0.5 cal · g^-1 · °C^-1, latent heat of fusion of ice = 80 cal · g^-1
Solution:

A copper calorimeter of mass 300 g contains 270 g of water at 30° C. Now 20 g of ice at -10°C is added to it.

Heat lost by calorimeter and water in cooling from 30°C to 0°C = 300 x 0.1 x 30 + 270 x 30 = 9000 cal

Heat taken by ice to rise from -10°C to 0°C = 20 x 0.5 x 10 = 100 cal

Heat taken for melting of ice = 20 x 80 = 1600 cal.

Total heat required to raise 20 g of ice from -10°C to 0°C and to melt it = 100 + 1600 = 1700 cal.

As heat lost is greater than heat required, the whole of ice will melt and the extra (9000 – 1700) or 7300 cal of heat will increase the temperature of the calorimeter and the mixture. Let the final temperature = t°C.

∴ 300 x 0.1 x t+ (270 + 20) x 1 x t = 7300

or, \(320 t=7300 \text { or, } t=\frac{7300}{320}=22.81^{\circ} \mathrm{C}\)

∴ Final temperature of the mixture = 22.81 °C.

Example 2. 90 g of water at 40°C is kept in a calorimeter of water equivalent lOg. What is the effect of immersion 100 g of ice at -10°C in it? The specific heat capacity of ice is = 0.5 cal · g^-1 °C^-1, latent heat of fusion of ice = 80 cal · g^-1.
Solution:

90 g of water at 40°C is kept in a calorimeter of water equivalent lOg.

Maximum heat that can be lost by the calorimeter and the contained water in cooling up to 0°C from 40°C = (10 + 90) x 40 = 4000 cal.

Heat taken by 100 g of ice to rise from -10°C to 0°C = 100 x 0.5 x 10 = 500 cal.
0°C

∴ Total heat required = 500 + 8000 = 8500 cal

Since, heat required is more than the maximum heat available, some ice will not melt. Let us assume that mg of ice melts.

∴ \(500+80 m=4000 \quad \text { or, } 80 m=3500\)

or, m = \(\frac{3500}{80}=43.75\)

Hence, only 43.75 g ice will melt. Mass of ice that remains unchanged = 100-43.75 = 56.25 g. The final temperature of the mixture will be 0°C.

Therefore, the calorimeter will finally contain 56.25 g of ice and (90 + 43.75) g or 133.75 g of water.

Example 3. Densities of ice and water at 0°C are 0.916 g- cm-3 and 1 g · cm^-3 respectively. A metal piece of mass 10 g at 100°C is dropped in a mixture of ice and water. Some ice melts and the volume of the mixture decreases by 0.1cm³ without any change in temperature. If latent heat of fusion of ice is 80 cal · g^-1, find the specific heat of the metal.
Solution:

Densities of ice and water at 0°C are 0.916 g- cm-3 and 1 g · cm^-3 respectively. A metal piece of mass 10 g at 100°C is dropped in a mixture of ice and water. Some ice melts and the volume of the mixture decreases by 0.1cm³ without any change in temperature. If latent heat of fusion of ice is 80 cal · g^-1

Volume of 1 g ice at 0°C = 1/0.916 = 1.092 cm³

Volume of 1g of water at 0°C = 1/1 = 1 cm³

Hence, contraction in volume on melting of 1 g of ice = 1.092- 1 = 0.092 cm³

∴ Contraction of 0.1 cm³ in volume is due lo melting of \(\frac{0.1}{0.092} \mathrm{~g}\) or 1.087 8 of ice.

Heat needed to melt 1.087 g of ice = 1.087 x 80 cal

Let specific heat of the metal be s.

⇔ 10 x s x 100 = 1.087 x 80

or, \(s=\frac{1.087 \times 80}{1000}=0.087 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} \text {. }\)

Example 4. A calorimeter of mass 50 g contains 200 g of water j at 20°C. 20 g ice at 0°C is mixed with it. Final temperature of the mixture becomes 11 °C. What is the latent heat of fusion of ice? The specific heat of the material of the calorimeter is 0.095 cal · g^-1 · °C^-1.
Solution:

A calorimeter of mass 50 g contains 200 g of water j at 20°C. 20 g ice at 0°C is mixed with it.

Let latent heat of fusion of ice = L cal • g^-1.

Heat taken to change 20 g of ice at 0°C to water at 0°C = 20 L cal

Heat taken by 20 g of water to rise to 11°C from 0°C = 20x 1 x 11 = 220 cal.

Heat lost by the calorimeter and its contents in cooling from 20°C to 11°C

= 50 x 0.095 x (20 – 11)+ 200 x 1 x (20-11)

= 1842.75 cal

Heat gain = heat lost

∴ 220 + 20L = 1842.75 or, L = \(\frac{1622.75}{20}=\) = 81.137

Therefore, the latent heat of fusion of ice is 81.137 cal · g^-1.

Example 5. When a solid of mass 50 g is heated at a constant rate of 5 cal · s^-1, there is a rise in temperature of 11°C in the first minute. For the next 13 min of heating, the temperature remains constant. Rate of rise of temperature after this is 6°C ·min^-1. Find

1. specific heat in the solid state,
2. specific heat in the liquid state and
3. latent heat of fusion of the solid.

Solution:

When a solid of mass 50 g is heated at a constant rate of 5 cal · s^-1, there is a rise in temperature of 11°C in the first minute. For the next 13 min of heating, the temperature remains constant. Rate of rise of temperature after this is 6°C ·min^-1.

1. Let specific heat in the solid state = s₁

As per data, 50 x s₁ x 11 = 5 x 60

or, s₁ = \(\frac{300}{550}\) = 0.545 cal · g^-1 · °C^-1.

2. The second phase of heating, is a case of fusion of the solid since the temperature remains constant.

Let latent heat of fusion = L

∴ 50 x L = 13 x 60 x 5

∴ L = 78 cal · g^-1.

3. In the third phase of heating, the temperature of the liquid state of the specimen increases.

Let s₂ = specific heat of the substance in its liquid state.

∴ 50 x s₂ x 6 = 5 x 60 or, s₂ = 1 cal · g^-1 · °C^-1.

Change of Volume During Melting and Solidification: The change from the liquid to the solid state and vice versa involves a change in volume, even though the temperature remains constant.

• For most materials, the volume decreases when a liquid changes to its solid state, and so the density increases. Conversely a change from solid to liquid state results in an increase in volume and therefore a decrease in density.
• But there are a few exceptions to this phenomenon—ice, cast iron, brass, antimony, bismuth etc., decrease in volume on melting and increase in volume on solidification. So, these objects have lower densities in their solid states.

Advantages and disadvantages of change in volume with the change of state: it is seen that the volume of water increases when it solidifies. This expansion may create some problems. In cold countries, the water inside the pipes may freeze into ice during winters.

• The expansion in volume exerts a huge force on the pipes causing them to burst. Again, hot water pipes burst more often than cold water pipes do. The amount of air dissolved in cold water is more than in hot water.
• So when cold water freezes the dissolved air comes out of the ice. If it is possible for air to move out of the pipe the ice can occupies the space of that air. But hot water has less dissolved air. So hot water pipes are more prone to burst compared to cold water pipes.
• In the same way, water deposits in the hill crevices expand on solidification and produce cracks on the stone surface.
• The expansion of volume on solidification has some advantages too. It facilitates the survival of fish and other marine creatures even in extreme cold weather.

Metals such as cast iron and brass expand on solidification. This property makes it suitable for casting in foundries. As molten iron solidifies inside the cast, it expands and fills up the volume evenly. The object conforms to the shape of the cast perfectly without creating any defects.

Effect of Pressure on Melting Point

1. For substances whose volume decreases on melting: The melting point of a substance whose volume decreases on melting, decreases with increase in pressure i.e. the substance melts at a lower temperature.

• Such substances are ice, brass, cast iron, antimony, bismuth etc. Increase in pressure helps in the decrease in volume. This makes melting easier and therefore the melting point decreases.
• It is found that the melting point of ice decreases by 0.0073°C when the pressure on it is increased by one standard atmosphere.

Regelation: Regelation is defined as melting of ice under application of pressure and resolidification of the molten ice on withdrawal of the pressure. When two pieces of ice at 0°C are pressed together, the melting point at the surfaces in contact falls below 0°C .

• But the temperature of ice remains 0°C; so ice melts at the junction. As the necessary latent heat is drawn from the ice itself, the temperature of the contact surface and of the molten ice falls below 0°C. On release of pressure, the melting point rises to 0°C again.
• Water at the surface of contact, being at a temperature lower than 0°C, freezes and the two pieces stick together to form a single piece of ice. This process of resolidification of water into ice is known as regelation.

2. For substances whose volume increases on melting: Substances like wax, copper, naphthalene etc. increase in volume on melting. An increase in pressure on them increases the melting point i.e. they melt at a higher temperature.

As increased pressure opposes expansion in volume, melting becomes difficult and thus melting point increases. One standard atmosphere rise in pressure raises the melting point of wax by 0.04°C.

Laws of Fusion: The facts related to melting can be summarised as the following laws. These laws are known as laws of fusion.

1. Any solid, at a fixed pressure, melts at a fixed tempera¬ture. The temperature remains constant until the whole solid specimen melts. This temperature is called the melting point of the solid at that pressure. Melting point of a solid is the same as tire freezing point of its liquid state. Different substances have different melting points.
2. For solids which decrease in volume on melting, the melting point decreases on increase in pressure. Increase in pressure, however, increases the melting point of those solids which expand on melting.
3. Unit mass of a solid absorbs a fixed quantity of heat during melting at a constant temperature. This quantity of heat is called its latent heat of fusion. Unit mass of the same material in liquid state releases the same amount of heat while freezing, The latent heat of fusion or latent heat of solidification is fixed for every substance but is different for different substances.
4. Freezing point of a solution is always less than the freez¬ing point of the pure solvent.
5. The melting point of an alloy is always less than the individual melting points of all of its constituent metals.

Unit 7 Properties Of Bulk Matter Chapter 8 Change Of State Of Matter Short Answer Type Questions

Question 1. Tea gets cooled when sugar is added to it. Why?
Answer:

Tea gets cooled when sugar is added to it.

When sugar is added to tea, it melts. For this purpose, the required latent heat is collected by sugar from tea, thereby cooling the tea.

Question 2. The water equivalent of a calorimeter is 10 g and it contains 50 g of water at 15 °C. Some amount of ice, initially at -10°C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice = 0.5 cal · g^-1 · °C^-1, specific heat of water = 1.0 cal · g^-1 · °C^-1 and latent heat of melting of ice = 80 cal · g^-1)?

1. 10g
2. 18 g
3. 20 g
4. 30 g

Answer:

Let, ice of m g is dropped in the calorimeter.

m x 0.5 x 10 + m/2 x 80 =60x 1 x 15

or, m = 20 g

The option 3 is correct.

Question 3. Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80 °C, the mass of water present will be [Take specific heat of water = 1 cal · g^-1 · °C^-1 and latent heat of steam = 540 cal · g^-1]

1. 24 g
2. 31.5 g
3. 42.5 g
4. 22.5 g

Answer:

Let m g of steam is passed into 20 g of water at 10° C.

Total heat released by the steam

= m x 540 + m( 100-80) =560/72 cal

Total heat absorbed by the water = 20(80-10) = 1400 cal

∴ 560m = 1400 or, m = 2.5 g

Now, mass of water present = 20 + 2.5 = 22.5 g.

The option 4 is correct.

Question 4. A piece of ice falls from a height h so that it melts completely, Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [latent heat of ice is 3.4 x 15 J/kg and g = 10N/kg]

1. 544 km
2. 136 km
3. 68 km
4. 34 km

Answer:

Heat produced = mgh [ m = mass of ice]

According to the question,

1/4 (mgh) = mL (L = latent heat of fusion of ice]

or, \(h=\frac{4 L}{g}=\frac{4 \times 3.4 \times 10^5}{10}=13.6 \times 10^4 \mathrm{~m}=136 \mathrm{~km}\)

The option 2 is correct.

Question 5. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (specific heat of copper = 0.39 J · g^-1 · K^-1; heat of fusion of water = 335 J · g^-1)
Answer:

The ice block is at 0°C.

So the heat released by the copper block = 2500 x 0.39 x (500 – 0) = 25 x 39 x 500 J

The maximum amount m of ice would melt if the entire amount of heat released by the copper block is used up as latent heat of fusion of ice.

∴ m = \(\frac{25 \times 39 \times 500}{335}\) = 1455 g =1.455 kg

Unit 7 Properties Of Bulk Matter Chapter 8 Change Of State Of Matter Integer Type Question And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. A piece of ice (heat capacity =2100 J • kg^-1 • °C^-1 and latent heat = 3.36 x 10⁵ J • kg^-1) of mass m g is at -5°C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally, when the ice-water mixture is in equilibrium, it is found that 1 g of ice has melted. Assuming there is no other heat exchange in the process, what will be the value of m?
Answer: 8

Question 2. 2 kg of ice at -20 °C is mixed with 5 kg of water 20 °C in an insulating vessel having a negligible heat capacity. Calculate the final mass (in kg) of water remaining in the container. It is given that the specific heat of water and ice are 1 kcal · kg^-1 · °C^-1 and 0.5 kcal · kg^-1 · °C^-1 respectively while the latent heat of fusion of ice is 80 kcal · kg^-1.
Answer: 6

Question 3. In an industrial process, 10 kg of water per hour is to be heated from 20°C to 80°C. To do this steam at 150°C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90°C. How many kg of steam is required per hour? (specific heat of steam = 1 cal · g^-1 · °C^-1, steam = 540 cal · g^-1)
Answer: 1

Question 4. A bullet of mass 10 x 10^-3 kg moving with a speed of 20 m • s^-1 hits an ice block (0°C) of 990 g kept at rest on a frictionless floor and gets embedded in it If ice takes 50% of KE lost by the system, the amount of ice melted (in grams) approximately is n x 10^-3. Find the value of n.(J = 4.2 J • cal^-1, latent heat of ice = 80 cal • g^-1)
Answer: 3

Unit 7 Properties Of Bulk Matter Chapter 8 Change Of State Of Matter Comprehension Type Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. An immersion heater, in an insulated vessel of negligible heat capacity brings 10 g of water to the boiling point from 16°C in 7 min. The water is replaced by 200 g of alcohol, which is heated from 16°C to the boiling point of 78 °C in 6 min 12 s, 30 g are vaporised in 5 min 6 s.

1. Power of heater is nearly

1. 8.4 x 10³ J · s^-1
2. 84 W
3. 8.4 x 10³ cal · s^-1
4. 20 W

Answer: 2. 84 W

2. The specific heat of alcohol is

1. 0.6 J · kg^-1 ·°C^-1
2. 0.6 cal · g^-1 · °C^-1
3. 0.6 cal · kg^-1· °C^-1
4. 0.6 J · °C^-1

Answer: 2. 0.6 cal · g^-1 · °C^-1

3. The latent heat of vaporization of alcohol is

1. 854 J · kg^-1
2. 854 x 10³ J · kg^-1
3. 204 cal · g^-1
4. 204 cal · kg^-1

Answer: 2. 854 x 10³ J · kg^-1

Question 2. In a physics practical examination, a student puts heat into a 500 g sample (solid) at the rate of 10 kJ · min^-1, while recording its temperature as a function of time. He collects the data and plots the graph depicted. Assume no heat is lost.

1. What is the latent heat of fusion?

1. 35 kJ · kg^-1
2. 70 kJ · kg^-1
3. 25 kJ · kg^-1
4. 30 kJ · kg^-1

Answer: 4. 30 kJ · kg^-1

2. What is the specific heat of the substance in liquid phase?

1. 1 kJ · kg^-1 · K^-1
2. 1.5 kJ · kg^-1 · K^-1
3. 6.67 kJ · kg^-1 · K^-1
4. None of these

Answer: 1. 1 kJ · kg^-1 · K^-1

3. What is the specific heat of the substance in solid phase?

1. \(\frac{4}{3} \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)
2. \(\frac{3}{4} \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)
3. \(\frac{40}{3} \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)
4. \(1 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

Answer: 1. \(\frac{4}{3} \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}\)

 

 

Unit 7 Properties Of Bulk Matter

Chapter 8 Change Of State Of Matter Long Answer Type Questions

Question 1. A solid is heated at a constant rate. Temperature of the specimen changes. Study the graph and answer the following questions.

1. What do the horizontal parts, AB and CD In the graph, represent?
2. If CD = 2AB, what inference can be drawn?
3. What does the slope of DE indicate?
4. The slope of OA is more than the slope of BC. What is the inference?

Answer:

1. Horizontal parts AB and CD represent respective quantities of heat absorbed for the changes of state at respective constant temperatures.

AB represents melting and CD represents the vaporisation of the substance.

2. CD = 2AB indicates that, latent heat of vaporisation is twice the latent heat of fusion, for the material in question.

3. From the slope of DE, the reciprocal of thermal capac¬ity of the specimen in vapour state can be calculated.

4. The slope of OA is greater than that of BC. It shows that the specific heat in liquid state is more than that in solid state.

Read and Learn More Class 11 Physics Long Answer Questions

Question 2. Why Is ice at 0°C more effective in cooling of a body than an equal amount of water at 0°C?
Answer:

Ice Is at 0°C more effective in cooling of a body than an equal amount of water at 0°C

When 80 cal of heat is extracted from 1 g of water at 0°C, it changes to 1 g of ice at 0°C. Hence, 1 g ice can absorb 80 cal more heat than 1 g water at 0°C. Thus, to cool a body, ice at 0°C is a more effective than an equal amount of water at 0°C.

Question 3. Water at 100 C and Meant of the same amount at 100 C— which one can release more heat? Which will cause more severe burns?
Answer:

Water at 100 C and Meant of the same amount at 100 C

An addition of 540 cal of heat to 1 g of water at J00°C produces steam at 100°C and therefore, the steam tan release more heat than water at 100°C. Thus, steam produces more severe burns than water of the same mass at 100°C.

Question 4. Some water is kept In a hole drilled into a melting Ice block. Will the water freeze to ice?
Answer:

Some water is kept In a hole drilled into a melting Ice block.

Water will not freeze to ice. The melting ice block is at 0°C. Water, if initially at a higher temperature, gives heat to the ice block at the hole, melting more ice.

Once water reaches 0°C, thermal equilibrium is established between the block and water and so heat exchange stops. The water can not transform into ice as it does not release any latent heat.

Question 5. State whether water in a beaker can be made to boil with a flow of steam at normal pressure.
Answer:

Water in the beaker will not boil. Steam is at 100°C under normal pressure. Steam condenses to water by delivering the latent heat of condensation and the water in the beaker absorbs it to rise up to 100°C.

Heat exchange stops at this stage. Hence, water in the beaker cannot take any latent heat of vaporisation from the steam. So, water will remain at 100°C but will not boil.

Question 6. An astronaut carried some water at 20°C in a thermoflask and on the moon he poured it into a beaker. Comment on what happens to the water.
Answer:

An astronaut carried some water at 20°C in a thermoflask and on the moon he poured it into a beaker.

Moon has no atmosphere and hence no air pressure acts there. With the lowering of pressure on a liquid surface, its boiling point decreases and the rate of vaporisation becomes very high.

Thus, water from the flask at 20°C starts boiling and vaporises very fast collecting the necessary latent heat from the rest of the mass. Hence, the temperature of the rest of the water decreases and ultimately it changes to ice in the beaker.

Question 7. Why is a cooling effect produced due to evaporation?
Answer:

When a liquid changes to its vapour state it requires latent heat. If no heat is supplied from outside then the liquid takes the required latent heat from its own body and from its surroundings. As a result, the liquid and its surroundings become cooler. So, a cooling effect is produced due to evaporation.

Question 8. Melting point of ice decreases on increasing the pressure on ice, but melting point of wax increases with the Increase in pressure. Explain with reasons.
Answer:

Melting point of ice decreases on increasing the pressure on ice, but melting point of wax increases with the Increase in pressure.

For solids which contract on melting, like ice, the melting point decreases on increase of the pressure on it (as increased pressure helps contraction). But for solids which expand on melting, like wax, an increase in pressure opposes the expansion and the melting point increases.

Question 9. Why does it take comparatively more time to completely convert some water to its vapour state than to attain boiling point from 0°C, by supplying heat continuously at the same rate?
Answer:

To convert 1 g of water from 0°C to 100°C, 100 cal of heat is required. But 1 g of water at boiling point takes 537 cal to change to its vapour state. So, time taken to change the state of 1 g of water at 100°C to its vapour form is 5.37 times the time taken to change the temperature of lg of water from 0°C to 100°C.

Question 10. Water boils at a lower temperature in vacuum and its temperature decreases during boiling. Why?
Answer:

Water boils at a lower temperature in vacuum and its temperature decreases during boiling.

We know that a decrease in the pressure on a liquid results in a decrease in its boiling point. So, water boils at a lower temperature in vacuum. If no heat is supplied externally then water takes the latent heat from the remaining water. That is why the temperature of boiling water decreases.

Question 11. What will happen if we pour a few drops of ether on the bulb of a thermometer? Also, what will be the consequence if the bulb of a thermometer is kept completely immersed in a bottle of ether?
Answer:

As ether is volatile, the drops of ether will vaporise fast on contact with air. During vaporisation, it will take the necessary latent heat from the bulb of the thermometer.So, the reading of the thermometer will drop.

• However, when the bulb is kept completely immersed in a bottle of ether, it does not come in contact with air.
• So, the ether in contact with the bulb does not evaporate, nor is any latent heat absorbed from the thermometer so the reading does not drop. The reading of the thermometer, in this case, gives the temperature of the ether in the bottle.

Question 12. What is the function of the energy supplied by latent heat of fusion or of vaporisation?
Answer:

Change of a solid to its liquid state involves breaking of the crystal structure of the solid. Latent heat of fusion provides the energy for this. Similarly, to change a liquid to its vapour state, intermolecular separations need to be so increased that there remains practically no attraction between the molecules the latent heat of vaporisation supplies the required energy for this.

Question 13. Why is a fuse wire in electrical lines made up of an alloy and not of pure metals?
Answer:

Melting point of an alloy is always less than the melting point of each of the constituent pure metals. So, a fuse wire made of an alloy can melt before the current in the circuit reaches a danger level and hence can break the circuit. Thus the other parts of the circuit remains safe.

Question 14. More work has to be done to vaporise a substance than to melt it. Explain.
Answer:

More work has to be done to vaporise a substance than to melt it.

In melting, the crystal structure of the molecules of a solid breaks down. For this, in most cases, the intermolecular separation changes very little. So, the work done against the intermolecular force is comparatively low. During vaporisation, the intermolecular separation increases by a lot more. So, more work has to be done against the intermolecular force.

Question 15. Boiling point of water at Darjeeling is usually less than 100°C. Explain why?
Answer:

Boiling point of water at Darjeeling is usually less than 100°C.

Air pressure decreases as altitude increases. Lower pressure on a liquid lowers its boiling point. Darjeeling being at a higher altitude, has a low air pressure and therefore water has a lower boiling point; so water begins to boil at a temperature below 100°C.

Question 16. Water boils at nearly 120°C in a pressure cooker. Why?
Answer:

Water boils at nearly 120°C in a pressure cooker.

Boiling point increases with the increase in pressure on a liquid. Pressure on a liquid in a pressure cooker rises to about 2 atm and so the boiling point of water at that pressure is nearly 120°C.

Question 17. At 100°C, during boiling, what is the specific heat of water?
Answer:

Heat gained during boiling (H) = latent heat of vaporisation

Rise in temperature t = 0

Now, H = mst

So, specific heat, s = \(\frac{H}{mt}\).

According to this equation, the value of specific heat is infinite.

Question 18. Why does a liquid evaporate even at a temperature below its boiling point?
Answer:

The air around any liquid has a tendency to become saturated with the vapour of the liquid. A certain mass of air has a capacity of holding a certain maximum amount of the vapour.

• When that maximum amount is present, air is said to be saturated with that vapour. But in most cases, the air surrounding a liquid is unsaturated. So, in order to saturate the surrounding air, molecules on the surface of a liquid continuously transform into vapour molecules that mix with air, whatever the temperature of the liquid be.
• This is the principle of evaporation of a liquid at all temperatures. However, when air attains saturation or near saturation, the rate of evaporation becomes negligibly small.
• That is why wet clothes do not dry up easily in rainy season. When the temperature of air becomes very low in winter, the atmosphere often becomes supersaturated with water vapour. Then the vapour turns back into water particles, thus forming dews and fogs.

Question 19. Sometimes why does a layer of ice stick to the heel of j a shoe during walking on ice?
Answer:

Under the pressure of the heel of a shoe on ice, its melting point decreases and ice melts. Now on raising the shoe above, the pressure on ice diminishes and its melting point increases. So water again freezes into ice which stick to ihe heel of the shoe.

Question 20. Liquid oxygen at 50 K is heated to 300 K at a constant pressure of 1 atm. The rate of heating is constant. Which of the following graphs represents the variation of temperature with time?

Answer:

Liquid oxygen will undergo a change of state when heated from 50K to 300K. During the change of state, the temperature will not change until all the liquid aw gen has changed to gaseous state.

After that, the temperature will increase because the rate of heating is constant. Hence the correct graph is (c).

Question 21. The melting point of ice in vacuum is 0.0073°C What do you mean by the statement?
Answer:

The melting point of ice in vacuum is 0.0073°C

Melting point of ice reduces due to application of pressure on it. It has been found that the melting point of ice reduces by 0.0073°C for one-atmosphere increase of pres¬sure. Conversely, if the pressure is decreased by one atmosphere, i.e., if the pressure is nil. then ice will melt at 0.0073°C instead of at 0°C. So it can be said that the melting point of ice in vacuum will be 0.0073°C.

Properties Of Bulk Matter

Expansion Of Solid And Liquids Short Question And Answers

Question 1. The liquid with a co-efficient of volume expansion (γ) is filled in a container of a material having the coefficient of linear expansion a. If the liquid overflows on heating, then

1. γ = 3α
2. γ > 3α
3. γ <3α
4. γ > 3α³

Answer:

Apparent coefficient of volume expansion, γ’= γ-3α

Now, γ’ is positive when the liquid overflows on heating.

So the condition is γ – 3α > 0 or, ϒ > 3α

The option 2 is correct.

Question 2. When a copper sphere is heated, which physical quantity of the sphere will show maximum percentage change?
Answer:

When a copper sphere is heated, its volume will show maximum change in percentage.

Question 3. A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If L, a and Y respectively denote the length of the rod, coefficient of linear thermal expansion and Young’s modulus of its material, then for an increase in temperature of the rod by ΔT, the longitudinal stress developed in the rod is

1. Inversely proportional to α
2. Inversely proportional to Y
3. Directly proportional to ΔT/Y
4. Independent of L

Answer:

Longitudinal thermal stress = YαΔT.

The option 4 is correct.

Question 4. A solid rectangular sheet has two different coefficients of linear expansion α₁ and α₂ along its length and breadth respectively. The coefficient of surface expansion is (for α₁ t<<1 1, α₂ t<<l)

1. \(\frac{\alpha_1+\alpha_2}{2}\)
2. \(2\left(\alpha_1+\alpha_2\right)\)
3. \(\frac{4 \alpha_1 \alpha_2}{\alpha_1+\alpha_2}\)
4. \(\alpha_1+\alpha_2\)

Answer:

Let, initial length and breadth of rectangular sheet are a₁ and b₁ respectively.

If the final length and breadth are a₂ and b₂ respectively in increase in temperature t° C, then

⇒ \(a_2=a_1\left(1+a_1 t\right) \text { and } b_2=b_1\left(1+\alpha_2 t\right)\)

∴ \(a_2 b_2=a_1 b_1\left(1+\alpha_1 t\right)\left(1+\alpha_2 t\right)\)

or, \(a_1 b_1(1+\beta t)=a_1 b_1\left\{1+\left(\alpha_1+\alpha_2\right) t+\alpha_1 \alpha_2 t^2\right\}\)

or, \(1+\beta t=1+\left(\alpha_1+\alpha_2\right) t\)

[neglecting the term \(\alpha_1 \alpha_2 t^2\)]

∴ \(\beta=\alpha_1+\alpha_2\)

The option 4 is correct

Question 5. The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100 °C is (For steel, Young’s modulus is 2 x 10¹¹ N • m^-2 and coefficient of linear expansion is 1.1 x 10^-5 K^-1)

1. 2.2x 10⁸ Pa
2. 2.2x 10⁹ Pa
3. 2.2 x 10⁷ Pa
4. 2.2 x 10⁶ Pa

Answer: As length is constant, strain =ΔL/L = αΔQ

Now, pressure = stress = Yx strain

= 2x 10¹¹ x 1.1 x 10^-5 x 100

= 2.2 x 10⁸ Pa

The option 1 is correct.

Question 6. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (α) of the metal of the pendulum shaft are respectively

1. 25°C and 1.85 x 10^-5 l °C
2. 60°C and 1.85 x 10^-4 l °C
3. 30°C and 1.85 x 10^-3 l°C
4. 55°C and 1.85 x 10^-2 l °C

Answer:

Increase or decrease in time for change in temperature ΔT,

⇒ \(\Delta t =\frac{1}{2} \alpha \Delta T t\)

∴ 12 = \(\frac{1}{2} \alpha\left(40-T_0\right) \times 1\)

T₀ = temperature at which clock gives accurate breading, t = 1 day]

and 4 = \(\frac{1}{2} \alpha\left(T_0-20\right) \times 1\)….(2)

(1)+(2) we get,

3 = \(\frac{40-T_0}{T_0-20} \quad \text { or, } 3 T_0-60=40-T_0\)

or, \(T_0=\frac{100}{4}=25^{\circ} \mathrm{C}\)

Putting the value of T₀ in equation (1) we get,

⇒ \(15 \alpha=\frac{24}{24 \times 60 \times 60}\)

or, \(\alpha=\frac{1}{15 \times 60 \times 60}=1.85 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

The option 1 is correct.

Question 7. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by:

1. \(3 P K \alpha\)
2. \(\frac{P}{3 \alpha K}\)
3. \(\frac{P}{\alpha K}\)
4. \(\frac{3 a}{P K}\)

Answer:

Bulk modulus \(K=\frac{P}{\left(\frac{\Delta V}{V}\right)}\)

∴ \(\frac{\Delta V}{V}=\frac{P}{K}[\Delta V= change in volume ]\)

If increase in temperature Δt brings the cube to its original size, then

∴ \(\Delta V=V \cdot \gamma \Delta t \quad \text { or, } \Delta t=\frac{\Delta V}{V} \cdot \frac{1}{\gamma}=\frac{\Delta V}{V} \cdot \frac{1}{3 \alpha}=\frac{P}{3 K \alpha}\)

The option 2 is correct.

Question 8. Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ – l₁) is maintained same at all temperatures, which one of the following relations holds good?

1. \(\alpha_1 l_2^2=\alpha_2 l_1^2\)
2. \(\alpha_1^2 l_2=\alpha_2^2 l_1\)
3. \(\alpha_1 l_1=\alpha_2 l_2\)
4. \(\alpha_1 l_2=\alpha_2 l_1\)

Answer:

Let, lengths of brass rod at temperatures t₁ and t₂ are l₂ and l’₂ respectively.

∴ \(l_1^{\prime}=l_1\left\{1+\alpha_1\left(t_2-t_1\right)\right\}\)

Again, let lengths of steel rod at temperatures t₁ and t₂ are l₂ and l’₂ respectively.

∴ \(l_2^{\prime}=l_2\left\{1+\alpha_2\left(t_2-t_1\right)\right\}\)

According to the question, \(l_2-l_1=l_2^{\prime}-l_1^{\prime}\)

or, \(l_2-l_1=l_2\left\{1+\alpha_2\left(t_2-t_1\right)\right\}-l_1\left\{1+\alpha_1\left(t_2-t_1\right)\right\}\)

or, \(l_2-l_1=l_2+l_2 a_2\left(t_2-t_1\right)-l_1-l_1 a_1\left(t_2-t_1\right)\)

or, \(l_2-l_1=\left(l_2-l_1\right)+\left(l_2-t_1\right)\left(l_2 a_2-l_1 a_1\right)\)

or, \(\left(t_2-t_1\right)\left(l_2 a_2-l_1 a_1\right)=0\)

as \(\left(t_2-t_1\right) \neq 0\)

∴ \(\left(l_2 \alpha_2-l_1 \alpha_1\right)=0 \quad or, l_1 \alpha_1=l_2 a_2\)

The option 3 is correct.

Question 9. Show that the coefficient of superficial expansion of a rectangular sheet of the solid is twice its coefficient of linear expansion.
Answer:

Let a₁,b₁ be the length and breadth, respectively, of the rectangular sheet at the initial temperature t₁.

Then, S₁ = a₁b₁ = initial area

Now, the temperature is raised to t₂, where t₂  – t₁ =t

The new values of the length and breadth are, respectively, a₂ and b₂; area S₂ = a₂b₂.

If α be the coefficient of linear expansion, then

⇒ \(a_2=a_1(1+\alpha t) \text { and } b_2=b_1(1+\alpha t)\)

∴ \(S_2=a_2 b_2=a_1 b_1(1+\alpha t)^2=S_1(1+2 \alpha t)\)

(neglecting α²t² as α<<1)

On the other hand, if β be the coefficient of superficial expansion, then

S₂ = S₁(1 +βt)

On comparison, we have β = 2α.