WBBSE Solutions For Class 9 Physical Science Chapter 6 Calorimetry

Chapter 6 Calorimetry Synopsis

Heat is a form of energy and exchange of heat is measurable. The branch of science which deals with the exchange of heat of a body, i.e., principles, methods or matters related to measurement of absorbed heat or emitted heat is known as calorimetry.

Specific Heat is the quantity of heat required to increase the temperature of unit mass of a substance by unit amount.

Units Of Specific Heat in CGS — cal • g-1 • °C-1 and in SI — J • kg-1 • K-1.

\(1 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}=\frac{4.2 \mathrm{~J}}{\frac{1}{1000} \mathrm{~kg} \cdot \mathrm{K}}\) = 4200 J • kg-1 • K-1

Thermal Capacity Or Heat Capacity is the quantity of heat needed to raise the temperature of a body by unity.

Units of thermal capacity in CGS —  cal/°C and in SI —  J/K.

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Water Equivalent of a body is the mass of water for which the increase in temperature is same as that for the body when the amount of heat supplied is the same for both.

Units Of Water Equivalent in CGS—g and in SI—kg

The Amount Of Heat (H) Absorbed Or Released by a body during a thermal exchange depends on

  1. Mass of the body (m),
  2. Change of temperature (t1 ~ t2) of the body,
  3. Specific heat of the material of the body (s); i.e, H = m • s • (t1 ~ t2)

If two bodies of different temperatures are kept in contact with each other, the hotter body emits, heat and the other one receives that heat. If there is no loss of heat by any means, then at the time of thermal equilibrium, heat emitted by the hotter body = heat received by the colder body This is the fundamental principle of calorimetry.

Units Of Heat in CGS system and SI are calorie (cal) and joule (J), respectively. [Here, 1 cal = 4.1855 J]

Chapter 6 Calorimetry Synopsis Short And Long Answer Type Questions

Question 1. What do you mean by calorimetry?

Answer:

Calorimetry:-

Heat is a kind of energy and exchange of heat of a body is measurable. Calorimetry is that branch of science which deals with the exchange of heat of a body, i.e., principles, methods, or matters related to measurement of absorbed heat or emitted heat.

Question 2. Explain the basic principle of calorimetry.

Answer:

Basic Principle Of Calorimetry:-

If two bodies with different temperatures are kept in contact with each other, the body with the higher temperature releases heat to become colder and the other one accepts this heat to become hotter.

If there is no loss of heat, then at thermal equilibrium, heat emitted by the hotter body = heat received by the colder body This is the basic principle of calorimetry.

Suppose two bodies of masses m1, m2, and respective temperatures t1, t2 (t1 > t2) are kept in contact with each other. The first body emits heat and the second one receives heat. Let us assume that at the time of thermal equilibrium, the common temperature is t.

If the specific heats of the first and the second body are s1 and s2 respectively, then heat emitted by the first body, Q1 = m1s1(t1 -t), and heat received by the second body, Q2 = m2s2(t – t2).

Now as per the basic principle of calorimetry,

Q1 = Q2 or, m1s1(t1 – t) = m2s2(t – t2)

Question 3. What are the conditions for the applicability of the basic principle of calorimetry?

Answer:

Conditions Applicable For Basic Principle Of Calorimetry:-

The conditions for applicability of the basic principle of calorimetry are:

  1. During exchange of heat among different bodies with varied temperature, there should not be any exchange of heat with the surroundings.
  2. There should not be any chemical reaction between the bodies.
  3. If there is any exchange of heat between a liquid and a solid body, then the solid body must not dissolve in the liquid.
  4. There should not be any kind of physical change in the bodies while absorbing or releasing heat.

Question 4. Write down the differences between heat and temperature.

Answer:

The differences between heat and temperature are given below

WBBSE Solutions For Class 9 Physical Science Chapter 6 Calorimetry Differences Between Heat And Temperature

Question 5. Two bodies of different temperatures are bought into thermal contact. What change in temperature of the two bodies will happen?

Answer:

When two bodies at different temperatures are bought in thermal contact, heat is transferred from the body at higher temperature to that at lower temperature. Therefore hotter body starts to cool down whereas the colder body starts to warm up.

The flow of heat continues till both reaches the same temperature. At thermal equilibrium temperature of both the bodies are equal.

Question 6. The temperature of three solid metallic bodies A, B, and C are 10°C, 20°C, and 10°C, respectively. If the three bodies are kept in contact with each other, which one releases heat and which one accepts?

Answer:

In this case, temperature of B is highest. So, B releases heat. The other two bodies A and C have the same temperature which is less than that of B. So, these two accept heat.

Question 7. What do you mean by specific heat of a substance?

Answer:

Specific Heat Of A Substance Means:-

Specific heat is the quantity of heat required to increase the temperature of unit mass of a substance by unit amount.

Question 8. What is thermal equilibrium? When several bodies are in thermal equilibrium which physical quantity is same for all of the bodies?

Answer:

Thermal Equilibrium

When several bodies at different temperatures are bought into thermal contact with each other, they will exchange heat among themselves and ultimately attain the same temperature. This state where all the bodies are at the same temperature is called thermal equilibrium, b When several bodies are at thermal equilibrium the physical quantity ‘temperature’ is same for all of the bodies.

Question 9. Why water is used in hot water bag?

Answer:

Water has a specific heat higher than everything (except ammonia). Hence a fixed mass of water gains more heat than any other liquid of the same mass for the same rise in temperature.

Consequently, it loses more heat during cooling. This heat is used for fomentation. So a hot water bottle remains effective for a long period of time. That is why water is used in hot water bag.

Question 10. What is water equivalent of a substance?

Answer:

The water equivalent of a body is the mass of water for which the increase in temperature is same as that for the body when the amount of heat supplied is the same for both.

Question 11. Write the factors on which the amount of heat absorbed or released by a body depends on.

Answer:

The amount of heat (H) absorbed or released by a body during a thermal exchange depends on

  1. Mass of the body (m),
  2. Change of temperature (t) of the body,
  3. Specific heat of the material of the body (s).

The relation among them is — H = m • s • t.

Question 12. Same amount of heat is supplied to two different bodies of equal mass but of different materials. The increase in temperature of the two bodies be equal or not?

Answer:

We know, given heat H = mst i.e., increase in temperature t = \(\frac{H}{m \cdot s}\).

Since for the two bodies, heat supplied (H) and mass (m) are equal but specific heat (s) of the material of the bodies are different, t ∝ 1/s.

Hence, the body with comparatively less specific heat will have a higher increase in temperature.

Question 13. Same amount of heat is supplied to equal mass of milk and water. Why milk warms quicker than water?

Answer:

Increase in temperature (t) of a substance is inversely proportional to specific heat (s) of the material of the substance when amount of heat (H) absorbed and mass of the body (m) are kept constant.

i.e., t ∝ 1/s, H, and m are constant

Specific heat of milk is less than that of the water. Hence milk warms quicker than water.

Question 14. Temperatures of two liquids of equal masses are t1 and t2 (t1 > t2) respectively. T is the final temperature of their mixture. Find the ratio of specific heat of the two liquids.

Answer:

As per the question, mass (m) of each liquid is same and specific heat of two liquids are s1 and s2 respectively.

According to basic principle of calorimetry, m • (t1 – t) • s1 = m • (t – t2) • s2

∴ \(\frac{s_1}{s_2}=\frac{t-t_2}{t_1-t}\)

or, s1 : s2 = (t – t2) : (t1 – t)

Question 15. What is the specific heat of water during its freezing at 0°C?

Answer:

Specific heat, s = \(\frac{H}{m \cdot t}\). During freezing, the temperature of water remains fixed at 0°C. So the change in temperature t = 0 hence h = ∞.

Chapter 6 Calorimetry Very Short Answer Type Questions Choose The Correct Answer

Question 1. If equal masses of two liquids at temperatures 10°C and 40°C are mixed together, the temperature of the mixture becomes 15°C. The ratio of specific heat of the two liquids is

  1. 5: 1
  2. 1: 5
  3. 2 : 3
  4. 4 : 3

Answer: 1. 5 : 1

Question 2. Heat is the of a material particle.

  1. Changed form of momentum
  2. Changed form of kinetic energy
  3. Changed form of potential energy
  4. Changed form of velocity

Answer: 2. Changed form of kinetic energy

Question 3. Thermometer is used to measure

  1. Temperature of a body
  2. Latent heat
  3. Radiated heat
  4. All of the above

Answer: 1. Temperature of a body

Question 4. Absorbed heat is _____ of a body.

  1. Directly proportional only to the mass
  2. Directly proportional only to the rise of temperature
  3. Directly proportional to the product of mass and increased temperature
  4. Directly proportional to the quotient of mass and increased temperature

Answer: 3. Directly proportional to the product of mass and increased temperature

Question 5. For calorimetric principle to be applicable

  1. Heat has to enter the body or the system from outside
  2. Heat has to go out of the body or the system
  3. A chemical reaction has to take place in the body or the system
  4. One body should not be soluble in another body

Answer: 4. One body should not be soluble in another body

Question 6. Among iron, mercury, water and air, which of the following has the highest specific heat?

  1. Water
  2. Iron
  3. Air
  4. Mercury

Answer: 1. Water

Question 7. A piece of iron and a wooden chair of the same mass are kept in sunlight. After keeping them for the same period of time, the piece of iron feels hotter on touching than the wooden chair due to

  1. Difference in water equivalent
  2. Difference in heat capacity
  3. Difference in heat conduction
  4. Difference in specific heat

Answer: 4. Difference in specific heat

Question 8. Heat accepted or rejected by a body is directly proportional to the increase or decrease of its temperature under the condition that

  1. Mass of the body is variable
  2. Mass of the body is fixed
  3. Water equivalent of the body is variable
  4. Latent heat of change of state of the body is constant

Answer: 2. Mass of the body is fixed

Question 9. According to the formula s = value of the specific heat of a body during melting becomes

  1. Zero
  2. 1
  3. Constant
  4. Infinite

Answer: 3. Constant

Question 10. According to the formula H = mst, which of the following is not true in case of a specific material?

  1. H ∝ m
  2. H ∝ t
  3. H ∝ s
  4. None of these

Answer: 3. H∝ s

Question 11. During fomentation, water bottle is used instead of iron. The main reason behind this is

  1. Specific heat of water is greater than that of iron
  2. Specific heat of iron is greater than that of water
  3. Iron gets heated slowly compared to water
  4. Water releases heat faster than iron

Answer: 1. Specific heat of water is greater than that of iron

Question 12. In the formula H = mst, t indicates

  1. The temperature of the body
  2. The change of temperature of the body
  3. The total time of heating
  4. All of the above

Answer: 2. The change of temperature of the body

Question 13. Mathematical form of the first law of thermodynamics is

  1. H = mL
  2. H = mst
  3. W = JH
  4. W = Fs

Answer: 3. W = JH

Question 14. Specific heat of copper is 0.1 cal • g-1 • °C-1. Water equivalent of a copper calorimeter of mass 0.4 kg is

  1. 40 g
  2. 4000 g
  3. 200 g
  4. 4 g

Answer: 1. 40 g

Question 15. What is the temperature difference between the top and bottom of a 100 m high waterfall, if the total amount of heat produced remains confined within the water?

  1. 0.23°C
  2. 0.46°C
  3. 0.15°C
  4. 0.69°C

Answer: 1. 0.23°C

Question 16. What is the final temperature of a mixture if 19.9 g of water at 30°C and 5g of ice at -20°C are mixed in a calorimeter? (specific heat of ice = 0.5 cal • g-1 • °C-1)

  1. 0°C
  2. -5°C
  3. 5°C
  4. 10°C

Answer: 3. 5°C

Question 17. Same masses of water and ice are mixed together at 0°C temperature. If the entire amount of ice melts, the initial temperature of water becomes

  1. 40°C
  2. 50°C
  3. 60°C
  4. 80°C

Answer: 4. 80°C

Question 18. Heat is supplied to a chunk of ice at the same rate. Ice starts melting after 5 s and in the next 40 s, the entire amount of ice melts. If the specific heat of ice is 0.5 cal • g-1 • °C-1, what is the initial temperature of ice?

  1. -20°C
  2. -22°C
  3. -15°C
  4. -10°C

Answer: 1. -20°C

Question 19. Calorimeters are generally made of

  1. Glass
  2. Copper
  3. Gold
  4. Wood

Answer: 2. Copper

Question 20. Which of the following substances has the highest specific heat?

  1. Water
  2. Alcohol
  3. Kerosene
  4. Milk

Answer: 1. Water

Question 21. cal • g-1 • °C-1 is the unit of

  1. Latent heat
  2. Water equivalent
  3. Thermal capacity
  4. Specific heat capacity

Answer: 4. Specific heat capacity

Question 22. Calorimetry relates to the measurement of

  1. Heat
  2. Temperature
  3. Mechanical energy
  4. Internal energy

Answer: 1. Heat

Question 23. Quantity of heat absorbed by a body depends on

  1. Mass
  2. Specific heat
  3. Increase in temperature
  4. All of them
  5. Answer: 4. All of them

Question 24. Thermal capacity of a body of mass m and specific heat s is

  1. \(\frac{m}{s}\)
  2. \(\frac{s}{m}\)
  3. ms
  4. \(\frac{1}{ms}\)

Answer: 3. ms

Question 25. Mean calorie means

  1. The amount of heat required to increase the temperature of 1 g of water from 0°C to 1°C
  2. The amount of heat required to increase the temperature of 1 g of water from 50°C to 51°C
  3. The amount of heat required to increase the temperature of 1 g of water from 14.5°C to 15.5°C
  4. 1/100 part of the amount of heat required 100 to increase the temperature of 1 g of water from 0°C to 100°C

Answer: 4. 1/100 part of the amount of heat required 100 to increase the temperature of 1 g of water from 0°C to 100°C

Question 26. During boiling of water at 100°C, what will be its specific heat?

  1. Zero
  2. 0.5
  3. 1
  4. Infinite

Answer: 4. Infinite

Question 27. Two bodies of different temperatures are mixed in a calorimeter. Which one of the following will remain conserved?

  1. Sum of the temperatures of the bodies
  2. Total heat of the bodies
  3. Total internal energies of the bodies
  4. Internal energy of each body

Answer: 3. Total internal energies of the bodies

Question 28. Which of the following pairs of physical quantities may have the same unit?

  1. Specific heat capacity and heat
  2. Heat capacity and water equivalent
  3. Specific heat capacity and heat capacity
  4. Heat and work

Answer: 4. Heat and work

Question 29. Which one of the following physical quantity is not relevant in calculation of heat absorbed by a body?

  1. Mass
  2. Density
  3. Specific heat capacity
  4. Increase in temperature

Answer: 2. Density

Question 30. Specific heat of lead in CGS is 0.03 cal • g-1 • °C-1. Specific heat of lead in SI is

  1. 12.6 J • kg-1 • K-1
  2. 126 J • kg-1 • K-1
  3. 1260 J • kg-1 • K-1
  4. 552 J • kg-1 • K-1

Answer: 2. 126 J • kg-1 • K-1

Question 31. Due to release of heat, electrical conductivity of a body

  1. Decreases
  2. Increases
  3. Remains constant
  4. Sometimes increases, sometimes decreases

Answer: 2. Increases

Chapter 6 Calorimetry Answer In Brief

Question 1. What is the unit of heat in CGS system?

Answer: Unit of heat in CGS system is calorie (cal)

Question 2. What is the unit of heat in SI?

Answer: Unit of heat in SI is joule (J).

Question 3. Boiled rice is prepared by boiling grain rice. What type of change takes place in this case due to absorption of heat?

Answer: Due to absorption of heat, chemical reaction takes place in this case.

Question 4. Heat is sometimes given to open the lid of stuck steel tiffin box. What is the role of heat in this case?

Answer: In this case, due to absorption of heat, pressure of gas inside the tiffin box increases, thereby opening the lid of the box.

Question 5. How is the heat received by a body related to the mass of the body?

Answer: For the increase of temperature, heat received by a body is directly proportional to the mass of the body.

Question 6. How is the heat received by a body of definite mass related to the temperature of the body?

Answer: Heat received by a body of definite mass is directly proportional to the increase of temperature of the body.

Question 7. Suppose the velocity of a meteorite of mass 52 kg is reduced from 15 km/s to 5 km/s while passing through the atmosphere of the earth. Calculate the amount of heat produced due to this change in velocity?

Answer: Amount of heat = 109 cal.

Question 8. What is the heat capacity of a body?

Answer: The amount of heat required to increase the temperature of the body by 1 kelvin (K) or 1°C is the heat capacity of the body.

Question 9. What is specific heat of a body?

Answer: Heat required to increase the temperature of a body by a unit (1°C or IK) is the specific heat of that body.

Question 10. If glucose is mixed with water, it gets completely dissolved. Is the basic principle of calorimetry applied here?

Answer: No, when one material is dissolved into another, basic principle of calorimetry is not applied.

Question 11. What does the quantity ms signify in the relationship, H = mst?

Answer: ms signifies heat capacity of the body.

Chapter 6 Calorimetry Fill In The Blanks

Question 1. Value of specific heat at the time of melting Of ice is __________

Answer: Infinte

Question 2. While ________ of heat of a body occurs, its temperature decreases.

Answer: Emission

Question 3. _________ of a body is defined as the amount of heat required to raise the temperature of the body by 1°C.

Answer: Heat capacity

Question 4. Unit of heat capacity in SI is _________

Answer: J.kg-1.K-1

Question 5. Specific heat of water in CGS system is taken as ________

Answer: 1

Question 6. Specific heat is sometimes referred to as _______ per unit mass of body.

Answer: Thermal capacity

Question 7. Specific heat of a substance can not be __________

Answer: Negative

Question 8. Two bodies are said to be in thermal equilibrium if they attain a _________ temperature.

Answer: Common

Chapter 6 Calorimetry State Whether True Or False

Question 1. According to Joule’s law, work done is directly proportional to the heat emitted.

Answer: True

Question 2. Specific heat of water is 4200 J • kg-1 • K-1.

Answer: True

Question 3. Temperature is the external manifestation of heat.

Answer: True

Question 4. Basic principle of calorimetry is not applicable for reaction of lime with water.

Answer: True

Question 5. The quantity of heat absorbed or given out by a body = weight of the body x specific heat capacity x change in temperature.

Answer: False

Question 6. Water equivalent indicates some volume of water.

Answer: False

Question 7. Water equivalent and heat capacity both can be measured by the term mL.

Answer: False

Question 8. Quantity of heat absorbed by a body depends on the medium of the surroundings.

Answer: False

Question 9. Thermal capacity and water equivalent of a body have some value in CGS system.

Answer: True

Chapter 6 Calorimetry Numerical Examples

Useful information

  1. Heat capacity of a body of mass m and specific heat s is = m • s.
  2. Quantity of heat absorbed or given out by a body (H) = mass of the body (m) x specific heat capacity (s) x change in temperature (t1 ~ t2).
  3. If two bodies of masses m1, m2; specific heat capacities s1, s2 and temperatures t1, t2 (t1 > t2) respectively are kept in thermal contact.
  4. At thermal equilibrium, final temperature is t. According to the basic principle of calorimeter, m1s1(t1-t) = m2s2(t-t2)

Question 1. Specific gravities of two liquids are 0.8 and 0.5, respectively. Equal amount of heat is required to raise the temperature of first liquid (3 L) and second liquid (2 L) by 1°C. What is the ratio of the specific heats of these two liquids?

Answer:

Specific gravities of two liquids are 0.8 and 0.5, respectively.

So, the densities of the liquids are 0.8 g/cm3 and 0.5 g/cm3, respectively.

Mass of 3 L or 3000 cm3 of first liquid,

m1 = 3000 x 0.8 = 2400 g

Mass of 2 L or 2000 cm3 of second liquid,

m2 = 2000 x 0.5 = 1000 g

Let us assume, s1 and s2 are the respective specific heats of first and second liquid.

According to the condition, we get

\(m_1 \times s_1 \times 1=m_2 \times s_2 \times 1\)

or, \(\frac{s_1}{s_2}=\frac{m_2}{m_1}\)

or, \(\frac{s_1}{s_2}=\frac{1000}{2400}\)

or, \(\frac{s_1}{s_2}=\frac{5}{12}\)

∴ \(s_1: s_2=5: 12\)

Hence, ratio of specific heats of two liquid is 5 : 12.

Question 2. A saucepan contains 100 g of water at 25°C. If 50 g of water of temperature 60°C is poured in it, the temperature of the mixture becomes 35°C. If the mass of the saucepan is 250 g, what is the specific heat of the constituent of the saucepan?

Answer:

Mass of water in sauce pan, m1 = 100 g

Initial temperature, t1 = 25°C

Mass of an extra amount of water poured in the pan, m2 = 50 g, and initial temperature, t2 = 60°C

Mass of the saucepan, m3 = 250 g

Final temperature of the mixture, t = 35°C

Specific heat of water, s1 = 1 cal • g-1 • °C-1

Now let us assume, specific heat of the constituent of the saucepan = s

So, heat absorbed by the saucepan and water of saucepan,

Q1 =m3 • s(t – t1) + m1 x s1 x (t – t1)

= 250 x s x (35 – 25) + 100 x 1 x (35-25)

=(2500s + 1000) cal

Heat released by water that is poured,

Q2 = m2 x s2 x (t2 -1)

= 50 x 1 x (60 – 25) = 1250 cal

From the basic principle of-calorimetry, we get Q1 = Q2

or, 2500s + 1000 = 1250 or, 2500s = 250

s = \(\frac{250}{2500}\) = 0.1 cal • g-1 • °C-1

Question 3. Temperature of a copper weight of 50 g is 90°C. What is the final temperature if the weight is immersed in 100 g water of temperature 10°C ? Specific heat of copper is 0.09 cal • g-1  • °C-1.

Answer:

Mass of the copper weight, m1 = 50 g

Initial temperature, t1 = 90°C

Specific heat, s1 = 0.09 cal • g-1 • °C-1

Mass of water, m2 = 100 g

Initial temperature, t2 = 10°C

Specific heat, s2 = 1 cal • g-1 • °C-1

Let us assume, final temperature = f.

So, heat released by the copper weight,

Q1 = m1s1(t1 – t)

= 50 x 0.09(90 -1) = 4.5(90 – t) cal

and heat absorbed by water,

Q2 = m2s2(t – t1)

= 100 x 1(t – 10) = 100 (t – 10) cal

From the basic principle of calorimetry, Q1 = Q2

or, 4.5(90 – t) = 100(t – 10)

or, 405 – 4.5t= 100t-1000

or, 104.5t = 1405

∴ t = \(\frac{1405}{104.5}\) = 13.44°c

Question 4. A platinum piece of mass 200 g is heated in a furnace and then slowly dipped in 650 g of water at 10° C in a vessel. This vessel has a mass of 500 g and specific heat of 0.1 cal • g-1 • °C-1. If the final temperature of the mixture is 25°C, what is the temperature of the furnace? Specific heat of platinum = 0.3 cal •  g-1 • °C-1.

Answer:

Suppose, temperature of the furnace is t1, then initial temperature of the platinum piece is also t1.

Mass of platinum, m1 = 200 g

Mass of water, m2 = 650 g

Initial temperature, t2 = 10°C

Specific heat, s2 = 1 cal • g-1 • °C-1

Mass of vessel, m3 = 500 g

Specific heat of constituent of vessel, s3 = 0.1 cal • g-1 • °C-1

Final temperature of mixture, t = 25°C

Now, heat released by the platinum piece,

Q1 = m1s1(t1 – t)

= 200 x 0.03 (t1 – 25) = (6t1 – 150) cal

and heat absorbed by vessel and water,

Q2 = m3s3(t – t2) + m2s2(t – t2)

= 500 x 0.1 x (25-10) + 650 x 1 x (25 – 10) = 10500 cal

From the basic principle of calorimetry, we get Q1 = Q2

or, 6t1 – 150 = 10500

∴ t = \(\frac{10650}{8}\)

Question 5. Temperature of three liquids of the same mass A, B, and C are 10°C, 30°C, and 4Q° C, respectively. If A and B are mixed together, the temperature of the mixture becomes 15°C. Again if B and C are mixed together, temperature of the mixture becomes 34°C. What is the ratio of specific heats of these three liquids?

Answer:

Suppose, specific heats of the three liquids A, B, and C are s1, s2, and s3, respectively and each liquid has a mass of m.

If A and B are mixed together, temperature of the mixture becomes 15°C.

From the basic principle of calorimetry, we get heat absorbed by liquid A = heat released by liquid B

∴ m x s1 x (15 – 10) = m x s2 x (30-15) or, 5s1 = 15s2

or, s2 = \(\frac{s_1}{3}\)

Again, if B and C are mixed together, temperature of the mixture becomes 34°C.

Again from basic principle of calorimetry, we get heat absorbed by liquid B = heat released by liquid C

∴ \(m \times s_2 \times(34-30)=m \times s_3 \times(40-34)\)

or, \(4 s_2=6 s_3 \quad \text { or, } s_3=\frac{2 s_2}{3}\)

or, \(s_3=\frac{2}{3} \times \frac{s_1}{3}=\frac{2 s_1}{9}\left[because s_2=\frac{s_1}{3}\right]\)

∴ ratio of specific heats of A , B, and C is

∴ \(s_1: s_2: s_3=s_1: \frac{s_1}{3}: \frac{2 s_1}{9}=9: 3: 2\)

Question 6. A 100 W powerful electric heater raises the temperature of 5 kg of a liquid from 25°C to 31°C in 2 minutes. What is the specific heat of the liquid?

Answer:

In 2 min = 60 x 2s = 120s, heat given by the electric heater,

Q = 100 x 120 J = 12000 J

Mass of the liquid, m = 5 kg

Increase in temperature,

t = (31 – 25)°C = 6°C

As we know, change of 1°C = change of 1 K

∴  increase in temperature, t = 6 K

Let specific heat of the liquid = s.

So, m x s x t = Q

or, 5 x s x 6 = 12000

∴ s = 400 J • kg-1 • K-1

Question 7. 90 g of water at 20°C is kept in a vessel of mass 100 g and specific heat of 0.1 cal • g-1 • °C-1. A piece of metal of mass 50 g is heated and then slowly dropped in this vessel. If decrease in temperature of the piece of metal is 20 times the increase in temperature of the water, then what is the specific heat of the constituent of the metal?

Answer:

Mass of the vessel, m1 = 100 g

Specific heat of the constituent of vessel, s1 = 0.1 cal • g-1  • °C-1

Mass of water, m2 = 90 g

Specific heat of water, s2 = 1 cal • g-1 • °C-1

Mass of metallic piece, m3 = 50 g

If increase in temperature of water is t, then according to the question, decrease of temperature of the piece of metal = 20t.

Now, heat absorbed by the vessel, and water is given by,

Q1 = m1 x s1 x t + m2 x s2 x t

= 100 x 0.1 x t + 90 x 1 x t

= 10t+ 90 = 100t cal

Let s be the specific heat of the constituent of the metal.

So, heat released by the metallic piece is given by

Q2 = m3 x s x 20t = 50 x s x 20t = 100t cal

From basic principle of calorimetry, Q1 =Q2

or, 100t = 1000st

∴ s = 0.1 cal • g-1 – °C-1

Question 8. Three liquids have masses m1, m2, and m3, temperature t1, t2, and t3 (t1 > t2 > t3) and specific heats s1 ,s2 and s3, respectively. If these three liquids are mixed together, what is the final temperature of the mixture?

Answer:

Suppose, final temperature of the mixture = t.

From basic principle of calorimetry,

\(m_1 s_1\left(t-t_1\right)+m_2 s_2\left(t-t_2\right)+m_3 s_3\left(t-t_3\right)=0\)

 

or, \(t\left(m_1 s_1+m_2 s_2+m_3 s_3\right)\)

= \(m_1 s_1 t_1+m_2 s_2 t_2+m_3 s_3 t_3\)

∴ \(t=\frac{m_1 s_1 t_1+m_2 s_2 t_2+m_3 s_3 t_3}{m_1 s_1+m_2 s_2+m_3 s_3}\)

Question 9. How much amount of water at 100°C is to be mixed with some amount of water at 20°C to make a mixture of volume 28 L at a temperature of 40°C?

Answer:

Suppose, volume of water at 100°C = x L and mass = 100x g

So, volume of water at 20°C = (28 – x)L and mass = (28 – x) x 1000 g

Specific heat of water = 1 cal • g-1 • °C-1

∴ heat absorbed by water at 20°C,

Q1 =(28 – x) x 1 x (40-20)

= 1000 x (28 – x) x 20 cal

and heat released by water at 100°C,

Q2 = 1000x x 1 x (100-40) = 1000x x 60 cal

From the basic principle of calorimetry, Q1 = Q2

or, (28 – x) x 1000 x 20 = l000x x 60 or, 28x – x = 3x or, 4x = 28

So, x = 7 and 28 – x = 28 – 7 = 21

Hence, 7 L of water at 100°C is to be mixed with 21 L of water at 20°C to make a mixture of volume 28 L at 40°C.

Question 10. Heat is applied to two different object of same mass at the same rate. 8 minutes and 10 minutes are required for raising the temperature by 20°C of the respective objects. Calculate the ratio of the specific heat of these two objects. If the specific heat of the first object is 0.2 cal • g-1 • °C-1, what is the specific heat of the second object?

Answer:

Let the mass of each object = m g.

Specific heat are s1 cal • g-1 • °C-1 and s2 cal • g-1 • °C-1, respectively.

Heat is applied to these objects at the rate of x cal/min.

So for the first object,

m x s1 x 20 = 8x ……(1)

and for the second object,

m x s2 x 20 = 10x ……(2)

Now by dividing equation (1) by equation (2), we get

\(\frac{m \times s_1 \times 20}{m \times s_2 \times 20}=\frac{8 x}{10 x}\)

 

or, \(\frac{s_1}{s_2}=\frac{4}{5}\)

So, the ratio of the specific heat of these two objects is s1: s2 = 4: 5.

Now, s1 = 0.2 cal • g-1 • °C-1.

Substituting the value in equation (3), we get

\(\frac{0.2}{s_2}=\frac{4}{5}\)

 

∴  s2 = 0.25 cal • g-1 • °C-1.

Question 11. Water from two taps are accumulating in a big vessel at the rate of 1.2 kg/min from the first and 0.8 kg/min from the second. Temperature of water from the first is 10°C and temperature from the second is 90°C. Now if both of these taps are opened simultaneously, then what is the temperature of accumulated water after 5 minutes? (Assume that there is no exchange of heat from the vessel.)

Answer:

After 5 minutes, mass of water emitted from the first tap, m1 = 1.2 x 5 = 6 kg

and mass of water emitted from the second tap, m2 =0.8 x 5 = 4 kg.

Let specific heat of water be s and final temperature of mixed water be t.

From the basic principle of calorimetry,

Q1 = Q2

or, m1 x s x (t – 10) = m2 x s x (90 -t)

or, 6(t – 10) = 4(90 – t) or, 6t – 60 = 360 – 4t

∴ t = 42°C

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