## Chapter 6 Equivalence Of Work And Heat Latent Heat Synopsis

- If work can be fully converted into heat, then work done and heat emitted are directly proportional to each other. This is Joule’s law.
- If H amount of heat is generated due to performance of work W, then W ∝ H or, W = JH; J being the mechanical equivalent of heat.
- Mechanical equivalent of heat is that amount of work, which is to be performed to generate one unit of heat. Mechanical equivalent of work, J = 4.2 x 10
^{7}erg/cal = 4.2 J/cal, Value of mechanical equivalent of heat in SI is 1. - A material may exist in different forms or physical states. These forms are physically distinct from each other but have the same chemical composition. By application of heat or extraction of heat or by any other mechanical method, forms of a material can be changed. Each form of the material is known as a distinct state or phase.

On the basis of the absorption or release of heat, change of state is of two types:

- Higher change of state and
- Lower change of state.

**Read and Learn More WBBSE Solutions for Class 9 Physical Science and Environment**

A change of state of a material due to absorption of heat is called higher change of state.

**Examples:** Melting or fusion, vaporisation.

On the other hand, a change of state of material due to release of heat is called lower change of state.

Example: solidification, condensation.

The quantity of heat released or absorbed per unit mass of a substance during its change from one state to another by keeping the temperature constant is called the latent heat of the substance.

Units of latent heat in CGS system and SI are cal/g and J/kg, respectively.

Here, 1 cal/g = 4200 J/kg.

Dimensional formula of latent heat is L^{2}T^{-2}.

## Chapter 6 Equivalence Of Work And Heat Latent Heat Short And Long Answer Type Questions

**Question 1. Write and explain Joule’s law regarding equivalence of work and heat.**

**Answer:**

**Joule’s Law Regarding Equivalence Of Work And Heat:-**

If work can be fully converted to heat, then work done and heat produced are directly proportional to each other. This is Joule’s law. Suppose, W amount of work is done to produce H amount of heat. According to Joule’s law,

W ∝ H or, W = JH

where J is a constant and is called mechanical equivalent of heat.

**Question 2. Define mechanical equivalent of heat. What is its value of erg/cal?**

**Answer:**

**Mechanical Equivalent Of Heat And Value Of erg/cal.**

- Mechanical equivalent of heat is defined as the amount of work that has to be done to produce one unit of heat.
- Value of J in CGS system is 4.2 x 10
^{7}erg/cal, this may be expressed as 4.2 J/cal.

**Question 3. Heat is produced when work is done give an example.**

**Answer:**

**Example For Heat Is Produced When Work Is Done:-**

When a wooden door or any material is polished with a piece of cloth by applying some pressure, we find that both our hand and the cloth warm up considerably. Polishing means reducing the roughness of the surface to make it more smooth. Work done against friction during polishing gets converted into heat.

**Question 4. Water in a waterfall falls from a height to the ground. Why is the temperature of water at the ground level slightly more than that at the top?**

**Answer:**

When water from the top of the waterfall falls towards the ground, its potential energy decreases. This decrease of potential energy is converted into kinetic energy. After water strikes the ground, a portion of this kinetic energy is converted to heat energy.

Now, a portion of this generated heat remains confined to water to increase its temperature. For this reason, temperature at the ground level is slightly higher than that at the top.

**Question 5. If a chunk of ice is dropped from a distant height, then why does a portion of it melt after striking the ground?**

**Answer:**

At normal temperature, if a chunk of ice at 0°C is dropped from a distant height to the ground, its potential energy decreases. This decrease of potential energy is converted into kinetic energy. After it strikes the ground, a portion of this kinetic energy is converted to heat energy.

Apart from this, heat is also generated due to its friction with air. Now a part of this produced heat remains confined to ice and melts a portion of the ice.

**Question 6. What do you mean by the state or phase of a material?**

**Answer:**

**State Or Phase Of A Material Means**

A material may exist in different forms. These forms are physically distinct from each other but have the same chemical composition. By absorption or release of heat or by any other mechanical means, the forms of a material can be changed. Each form of the material is known as a distinct state or phase.

**Question 7. What is the change of state of a material? How many types of change are envisaged on the basis of absorption or release of heat? What are those changes?**

**Answer:**

Change of state of a material is defined as the phenomenon of transformation of a material from one state to another due to either absorption or release of definite amount of heat to or from a material.

Two types of change are envisaged. The changes are:

- Higher change of state and
- Lower change of state.

**Question 8. What do you mean by higher change of state?**

**Answer:**

**Higher Change Of State Means:-**

A change of state of a material due to application of heat is called higher change of state.

A solid material can be converted to a liquid material and a liquid material can be converted to a gaseous material by absorption of heat. Transformation from solid to liquid state is called melting and transformation from liquid to gaseous state is called vaporisation. Thus, melting and vaporisation are higher change of state.

**Question 9. What do you mean by lower change of state?**

**Answer:**

**Lower Change Of State Means:-**

A change of state of a material due to release of heat is called lower change of state.

A liquid material can be converted to a solid material and a gaseous material can be converted to a liquid material by release of heat. Transformation from liquid to solid state is called solidification and transformation from gaseous to liquid state is called condensation. Thus, solidification and condensation are lower change of state.

**Question 10. With the help of an experiment, demonstrate that temperature remains constant during the time of change of state.**

**Answer:**

Some amount of ice is taken in a vessel. Now a thermometer is clamped by its side and its bulb is immersed in ice. Reading of the thermometer is now 0°C. Now the vessel is heated slowly. Ice starts melting due to absorption of heat.

It is observed that till the total amount of ice melts away, the reading of the thermometer remains constant at 0°C. After the total amount of ice melts in the water, it is observed that the reading of the thermometer starts increasing. It is inferred from this experiment that the temperature remains constant during the time of change of state.

**Question 11. What do you mean by latent heat?**

**Answer:**

**Latent Heat:-**

During the period of change of state of any material, whether there is any absorption or release of heat, there is not any change of temperature. The amount of heat released or absorbed per unit mass of a substance during the change from one state to another by keeping the pressure constant is called the latent heat of the substance for the constant change of state.

**Question 12. What are the units of latent heat in the CGS system and SI? Establish a relationship between these two units.**

**Answer:**

- Units of latent heat in the CGS system and SI are cal/g and J/kg, respectively.
- 1 cal/g = \(\frac{4.2 \mathrm{~J}}{10^{-3} \mathrm{~kg}}\) = 4200 J/kg

**Question 13. Calculate the dimensional formula of latent heat.**

**Answer:**

**Dimensional Formula Of Latent Heat:-**

= \(\frac{\text { dimenșional formula of heat }}{\text { dimensional formula of mass }}\)

= \(\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{M}}=\mathrm{L}^2 \mathrm{~T}^{-2}\)

**Question 14. What do you understand by the statement: Latent heat of melting of ice is 80 cal/g**

**Answer:**

The latent heat of melting of ice is 80 cal/g means that under normal pressure, 80 cal of heat is required to convert 1 g of ice at 0°C to 1 g of water at 0°C.

**Question 15. What do you understand by the statement: The latent heat of vaporization of water is 537 cal/g.**

**Answer:**

The latent heat of vaporization of water is 537 cal/g means that under normal pressure, 537 cal of heat is required to convert lg of water at 100°C to 1 g of vapor at 100°C.

**Question 16. Give a graphical explanation of latent heat.**

**Answer:**

**Graphical Explanation Of Latent Heat:-**

Solid material is taken. Suppose the initial temperature of the material is t_{1}. Point P indicates the initial temperature of the material. If heat is supplied to the material at a definite rate, its temperature increases.

This is represented by the line PQ in the graph. Then the material starts melting. Point Q indicates the beginning of melting. The temperature of point Q is the melting temperature t_{m}. Now, if heat is given continuously, the material melts and till the melting is complete, temperature remains constant.

Line QR in the graph represents the melting of the material. Melting is complete at point R where the entire solid material has been converted to liquid. Now if further heat is given, the temperature of the liquid increases. Line RS in the graph represents this increase.

In the same way, we may take a liquid material with an initial temperature of t_{2} and start releasing heat at a definite rate. Change of temperature with time is depicted. where point T denotes the initial temperature of the liquid.

The reduction of the temperature of the liquid is expressed by line TU. At point U, the solidification of liquid starts and it ends at point V.

The temperature remains unchanged from U to V. At point V, the entire liquid material has been converted to solid material. After this, if heat release goes on, the temperature of the solid is decreased which is represented by line VW.

QR portion and UV portion represent the latent heat of absorption and latent heat of release, respectively. These portions of the lines are straight lines parallel to the time axis because in these two cases, the temperature remains unchanged

**Question 17. Why does ice at 0°C feel colder than water at 0°C?**

**Answer:**

**Ice At 0°C Feel Colder Than Water At 0°C:-**

Under normal pressure; 1 g of ice at 0°C is obtained after the release of 80 cal of heat from 1 g of water at 0°C. This means that 1 g of ice at 0°C contains 80 cal less heat than 1 g of water at 0°C. Therefore, ice at 0°C feels colder.

**Question 18. Why does steam at 100°C feel warmer than water at 100°C?**

**Answer:**

**Steam At 100°C Feel Warmer Than Water At 100°C:-**

Under normal pressure, 1 g of steam at 100°C is obtained after applying 537 cal of heat to lg of water at 100°C. This means that lg of Steam at 100°C contains 537 cal more heat than lg of water at 100°C. For this reason, steam at 100°C feels warmer than water at 100°C.

**Question 19. Why is the latent heat of vaporization higher than the latent heat of melting?**

**Answer:**

**Yes, Latent Heat Of Vaporisation is Higher Than the Latent Heat Of Melting:-**

There are strong intermolecular forces working amongst the molecules in a solid matter. Compared to these, intermolecular forces in liquid are less, and in gases, these forces are significant. When converted from liquid to gaseous state, intermolecular distances between the molecules increase more than when it is converted from solid to liquid state.

As a result, during the transformation from liquid to gaseous state, more work is to be done against these intermolecular attractive forces, requiring more heat. For this reason, the latent heat of vaporization is higher than the latent heat of melting.

**Question 20. Why is water kept in an ice tray at 0°C not converted into ice?**

**Answer:**

Water is kept in an ice tray. If the temperature of the water is more than 0°C, then that water releases heat to become water at 0°C and some amount of ice melts in the process. At this stage, as the temperature of ice and water are at 0°C, there is no exchange of heat. So, when water is kept in an ice tray, it does not get converted to ice.

## Chapter 6 Equivalence Of Work And Heat Latent Heat Very Short Answer Type Questions Choose The Correct Answer

Question 1. The amount of heat required to convert 100 g of water at a temperature of 100°C to 100 g of steam at a temperature of 100°C is (the latent heat of vaporization = 540 cal/g)

- 5400 cal
- 540000 cal
- 27000 cal
- 54000 cal

Answer: 4. 54000 cal

Question 2. If the mechanical equivalent of heat, J = 4.2 J/cal, how much work has to be performed to produce 5 cal of heat?

- 20J
- 42J
- 21J
- 8.4J

Answer: 3. 21J

Question 3. During the time of higher change of state

- Heat is absorbed
- Heat is released
- Heat is neither applied nor released
- Temperature increases

Answer: 1. Heat is absorbed

Question 4. During the time of lower change of state

- Heat is absorbed
- Heat is released
- Heat is neither absorbed nor released
- Temperature decreases

Answer: 2. Heat is released

Question 5. The value of the mechanical equivalent of heat in the CGS system is

- 4.2 J/cal
- 4.2 x 10
^{7}erg/cal - 4.2 x 10
^{6}erg/cal - 4.2 x 10
^{5}erg/cal

Answer: 2. 4.2 x 10^{7} erg/cal

Question 6. If the latent heat of vaporization of water is 540 cal/g, then its value in SI is

- 22.68 x 10
^{5}J/kg - 22.68 x 10
^{6}J/kg - 5.4 x 10
^{6}J/kg - 5.4 x 10
^{5}J/kg

Answer: 1. 22.68 x 10^{5} J/kg

Question 7. The correct relationship between work (W) and heat (H) is

- W = \(\frac{H}{J}\)
- \(\frac{W}{H}=\frac{1}{J^2}\)
- H = \(\frac{W}{J}\)
- W = \(\frac{J}{H}\)

Answer: 3. H = \(\frac{W}{J}\)

Question 8. At any point in time during the absorption of heat

- Only change of state takes place
- Only a change of temperature takes place
- Change of state or temperature takes place
- Both change of state and temperature take place

Answer: 3. Change of state or temperature takes place

Question 9. Due to the release of heat, the electrical conductivity of a body

- Decreases
- Increases
- Remains constant
- Sometimes increases, sometimes decreases

Answer: 3. Remains constant

Question 10. According to scientists Rumford and Davy, which energy is converted to heat energy?

- Chemical energy
- Electrical energy
- Mechanical energy
- Potential energy

Answer: 2. Electrical energy

Question 11. When mechanical energy is converted to heat energy, then

- the kinetic energy of the body increases
- The kinetic energy of atoms and molecules of the body increases
- The water equivalent of the body increases
- Specific heat of the body increases

Answer: 2. Kinetic energy of atoms and molecules of the body increases

Question 12. The amount of heat required to melt the ice of 1 g into the water at 0°C is

- 100 cal
- 500 cal
- 4 cal
- 80 cal

Answer: 4. 80 cal

Question 13. The amount of heat required to convert 1 g of water into vapour at 100°C is

- 537 cal
- 100 cal
- 80 cal
- 4 cal

Answer: 1. 537 cal

Question 14. On the upper surface of an uncovered liquid

- Condensation = vaporisation
- Condensation > vaporisation
- Condensation < vaporisation
- Data incomplete

Answer: 3. Condensation < vaporisation

Question 15. If there is no change of state, then which of the following quantities is unnecessary for the calculation of absorption or release of heat?

- Mass
- Latent heat
- Specific heat
- Change of temperature

Answer: 2. Latent heat

Question 16. If heat is applied to dry ice (solid carbon dioxide), then it is directly converted to gaseous carbon dioxide. This phenomenon is called

- Vaporization
- Boiling
- Sublimation
- Melting

Answer: 3. Sublimation

Question 17. If heat is applied to water at 0°C, then

- Density increases at first and then decreases
- Volume increases at first and then decreases
- Volume increases continuously
- Mass increases at first and then decreases

Answer: 1. Density increases at first and then decreases

Question 18. A piece of ice is floating in a glass of water at 3°C. If this ice melts completely, the level of water in the glass

- Decreases
- Increases
- Remains same
- Is unpredictable

Answer: 2. Increases

Question 19. A piece of ice is floating in a glass of water at 5°C. Melting of the piece of ice decreases the temperature of water to 4°C. At this condition, the level of water

- Decreases
- Increases
- Remains same
- Data incomplete

Answer: 1. Decreases

Question 20. On the upper surface of a liquid kept in a closed vessel

- Rate of condensation > rate of vaporization
- Rate of condensation < rate of vaporization
- Rate of condensation = rate of vaporization
- Rate of condensation ≥ rate of vaporization

Answer: 3. Rate of condensation = rate of vaporization

Question 21. Vaporization of water takes place

- At 0°C
- At 15°C
- Only at 100°C
- At any temperature from 0°C to 100°C

Answer: 4. At any temperature from 0°C to 100°C

Question 22. At what temperature latent heat is required for the conversion of water to vapor?

- At 0°C
- At 14.5°C
- Only at 100°C
- At a temperature between 0°C and 100°C

Answer: 4. At a temperature between 0°C and 100°C

Question 23. A phenomenon or state that does not depend on the amount of water vapour present in air is

- Falling of dew
- Formation of fog
- Freezing of water to form ice
- Relative humidity

Answer: 3. Freezing of water to form ice

Question 24. Latent heat of vaporization of water

- Depends on the amount of heat absorbed
- Depends on the mass of water
- Depends on the volume of water
- Does not depend on anything

Answer: 4. Does not depend on anything

Question 25. It takes 10 minutes to raise the temperature of some amount of water from 0°C to 100°C and to vaporize the same amount, it takes another 55 minutes with the help of a heater. What is the latent heat of vaporization in this case?

- 530 cal/g
- 540 cal/g
- 550 cal/g
- 560 cal/g

Answer: 3. 550 cal/g

Question 26. 30 g of steam at temperature 100°C is supplied to 100 g of water at 20°C. How much gram of steam is condensed due to this process? [Latent heat of condensation of steam =540 cal/g]

- 17.8 g
- 16.8 g
- 15.8 g
- 14.8 g

Answer: 4. 14.8 g

Question 27. Which of the following is a higher change of state?

- Solidification
- Vaporization
- Condensation
- None of these

Answer: 2. Vaporisation

Question 28. Which of the following is a lower change of state?

- Melting
- Vaporization
- Condensation
- None of these

Answer: 3. Condensation

Question 29. The dimensional formula of latent heat is

- LT
^{-2} - MT
^{-2} - L
^{2}T^{-2} - ML
^{2}T^{-2}

Answer: 3. L2T“2

Question 30. A lead bullet speeding with a velocity of 200 m/s hits a wall. If 25% kinetic energy of the bullet is converted into heat, what is the increase in temperature of lead? (specific heat of lead = 0.03 cal • g^{-1} • °C^{-1})

- 29.67°C
- 39.67°C
- 49.67°C
- 59.67°C

Answer: 2. 39.67°C

## Chapter 6 Equivalence Of Work And Heat Latent Heat Answer In Brief

**Question 1. Some amount of heat is required for a change of state of a material. What is that heat called?**

**Answer:** The heat that is required for the change of state of a material is called latent heat.

**Question 2. What is the unit of latent heat in the CGS system?**

**Answer:** The unit of latent heat in the CGS system is cal/g.

**Question 3. What is the unit of latent heat in SI?**

**Answer:** The unit of latent heat in SI is J/kg.

**Question 4. What type of change of state is represented by melting and vaporization?**

**Answer:** Melting and vaporization represent a higher change of state.

**Question 5. What type of change of state is represented by solidification and condensation?**

**Answer:** Solidification and condensation represent a lower change of state.

**Question 6. What is the value of the mechanical equivalent of heat in a CGS system?**

**Answer:** The value of the mechanical equivalent of heat in the CGS system is 4.2 x 107 erg/cal.

**Question 7. What is the value of the mechanical equivalent of heat in SI?**

**Answer:** The mechanical equivalent of heat in SI is 1.

**Question 8. If there is no change of temperature of a body in spite of heating, can there be a change of state of the body?**

**Answer:** Yes, there may not be a change in the temperature of a body but due to the application of heat, there may be a change in the state of the body.

**Question 9. What is sublimation?**

**Answer:** Sublimation is the process by which a solid material is converted directly to its gaseous state due to the absorption of heat.

**Question 10. What is the water equivalent of a body if its heat capacity is 50 cal/°C?**

**Answer:** The water equivalent of the body is 50g.

**Question 11. If work done is fully converted to heat, what is the relationship between work done and heat produced?**

**Answer:** Work done and heat produced are directly proportional to each other.

**Question 12. Name four fundamental states of matter.**

**Answer:** The four fundamental states of matter are solid, liquid, gaseous, and plasma state.

**Question 13. Write the unit of J in SI.**

**Answer:** J has no unit in SI.

**Question 14. What amount of heat will produce if 1J of work is fully converted into heat?**

**Answer:** The amount of heat produced = 1/4.2 = 0.238 cal.

**Question 15. To produce 1 kcal heat how much work is needed?**

**Answer:** The amount of work needed to produce 1 kcal of heat is = 1000 x 4.2 J = 4200 J.

**Question 16. What is the value of latent heat of fusion of ice in the CGS system?**

**Answer:** The value of latent heat of fusion of ice in the CGS system is 80 cal/g.

**Question 17. What is the latent heat of vaporization of water in cal • g ^{-1}?**

**Answer:** The value of latent heat of vaporization of water in cal • g^{-1} unit is 540.

**Question 18. What is the specific heat of water during boiling at 100°C?**

**Answer:** During boiling at 100°C specific heat of water is infinite.

**Question 19. What amount of heat is needed to convert 100 g of water at 100°C to steam at 100°C?**

**Answer:** The latent heat of vaporization of water is 537 cal/g.

∴ The amount of heat required = 537 x 100 = 53700 cal.

**Question 20. What is the latent heat of condensation?**

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

**Question 21. Why steam at 100°C is a better warming agent than water at 100°C?**

**Answer:** 1 g of steam at 100°C can transfer 537 cal more heat than 1 g of water at 100°C. That is why steam at 100°C is a better warming agent than water at 100°C

**Question 22. Under what conditions can heat be supplied to a body without causing any change in temperature?**

**Answer:** During the change from one state to another the substance absorbs heat without any change. in the temperature.

## Chapter 6 Equivalence Of Work And Heat Latent Heat Fill in The Blanks

Question 1. _______ is the amount of work that needs to be done to produce one unit of heat.

Answer: Mechanical equivalent of heat

Question 2. Dimension of L in the dimensional formula of latent heat is ________

Answer: 2

Question 3. On the basis of absorption or release of heat, change of state of any material is of _______ types.

Answer: Two

Question 4. Water falls from top to bottom in a waterfall. Compared to the water at the top, the temperature is slightly __________ at the bottom.

Answer: Higher

Question 5. Water kept in the ice tray at 0°C temperature _________ convert into ice.

Answer: Does not

Question 6. Matter requires latent heat for any change Of ________

Answer: State

Question 7. Latent heat of melting of ice is _________

Answer: 80 cal/g

Question 8. An amount of 8.4 J work has to be done to produce ________ of heat.

Answer: 2 cal

Question 9. _______ of liquid takes place at any temperature.

Answer: Vaporisation

Question 10. Rate of vaporization _________ with increase of temperature of liquid.

Answer: Increases

Question 11. Unit of water equivalent in SI is __________

Answer: kg

## Chapter 6 Equivalence Of Work And Heat Latent Heat State Whether True Or False

Question 1. The latent heat of melting is greater than the latent heat of vaporization.

Answer: True

Question 2. Heat is the transformed state of the kinetic energy of a particle.

Answer: True

Question 3. In the steam engine heat is converted into mechanical work.

Answer: True

Question 4. Steam at 100°C causes more severe burns than water at 100 °C.

Answer: True

Question 5. The gradient of temperature vs time graph of a substance has – ve value when heat is supplied to it.

Answer: False

Question 6. The gradient of temperature vs time graph of a substance has – ve value when heat is extracted from it.

Answer: True

Question 7. Sublimation is the direct conversion from solid to gaseous state.

Answer: True

## Chapter 6 Equivalence Of Work And Heat Latent Heat Numerical Examples

**Useful information**

If W amount of work is fully converted into heat and H is the amount of heat produced, then according to Joule’s law.

W = JH, J is a constant, called the mechanical equivalent of heat.

In CGS J = 4.2 x 10^{7} erg/cal and in SI J = 1 and J = 4.2 J/cal

If Q amount of heat to be supplied or given out to change the state of mass m of a substance, then, Q = mL or, L = \(\frac{Q}{m}\)

L = Latent heat of the substance

If a piece of ice of mass m hits the earth’s surface from a height h, if total kinetic energy is converted into heat the amount of energy released H = \(\frac{mgh}{J}\)

g = acceleration due to gravity.

**Question 1. what is the amount of work done to produce heat of 50 cal?**

**Answer:**

Heat produced, H = 50 cal

As the mechanical equivalent of heat, J = 4.2 J/cal, the amount of work done is given by W = JH = 4.2 x 50 = 210 J

**Question 2. A block is dragged 4 m on a rough floor. If the frictional force between the block and the floor is 40 N and the work done is fully transformed into heat, then how much heat is produced? (Mechanical equivalent of heat, J = 4.2 J/cal)**

**Answer:**

The frictional force between the block and the floor, F = 40 N

Displacement of the block, s = 4 m

Work done against friction,

W = Fs = 40 N x 4 m = 160 J

Now if the work done against friction is fully converted into heat, then the heat produced is given by

H = \(\frac{W}{J}=\frac{160}{4.2}\) = 38.1 cal

**Question 3. The temperature of a piece of lead is 27°C. What is the minimum velocity with which it should strike a wall so that the heat point of lead is 327°C, the specific heat of lead is 0.03 cal • g ^{-1} • °C^{-1}, latent heat of melting is 5 cal/g and the mechanical equivalent of heat, J = 4.2 J/cal.)**

**Answer:**

Suppose, the mass of the piece of lead = m g and its minimum velocity = v m/s

So the kinetic energy of the piece of lead,

E = 1/2 mv^{2} erg

Mechanical equivalent of heat,

J = 4.2 J/cal = 4.2 x 10^{7} erg/cal

Now if the kinetic energy of lead is fully transformed into heat after it strikes a wall, then the heat produced is given by

H = \(H=\frac{E}{J}=\frac{m v^2}{2 \times 4.2 \times 10^7} \mathrm{cal}\)

To melt the piece of lead, its temperature is to be raised from 27°C to 327°C.

So the heat required to increase the temperature is given by H_{1} = m x 0.03 x(327 – 27) = 9m cal

and the required latent heat for melting the piece of lead, H_{2} = m x 5 = 5m cal

According to the condition, H =H_{1} + H_{2}

or, \(\frac{m v^2}{2 \times 4.2 \times 10^7}=9 m+5 m=14 m\)

or, \(v^2=14 \times 2 \times 4.2 \times 10^7=11.76 \times 1-10^8\)

v = \(\sqrt{11.76 \times 10^3}=3.43 \times 10^4 \mathrm{~cm} / \mathrm{s}\)

**Question 4. Calculate the amount of work to be done to convert 100 g of ice at 0°C to water of temperature 100°C. (Latent heat of melting of ice =80 cal/g, the mechanical equivalent of heat, J = 4.2 J/cal.)**

**Answer:**

Latent heat required for melting of ice, H_{1} = 100 x 80 = 8000 cal

Necessary heat is required for raising 100 g of water at 0°C to a temperature of 100°C,

H_{2} = 100 x 1 x(100 – 0) = 10000 cal

So the heat required to convert 100 g of ice at 0°C to water of temperature 100°C is given by H =H_{1} + H_{2}

= 8000 + 10000 = 18000 cal

Hence the required work to be done,

W = JH = 4.2 X 18000 = 75600 J

**Question 5. What is the height from which a chunk of ice at 0°C has to be dropped so that it melts completely due to its impact on the ground?**

**Answer:**

Suppose the mass of the chunk of ice = m g and h is the height from where it has been dropped.

Now, just before the moment it strikes the ground, its kinetic energy is equal to the loss of potential energy = mg.

This kinetic energy is transformed into heat energy due to its impact on the ground.

Now if J is the mechanical equivalent of heat, then heat produced, H = \(\frac{mgh}{J}\).

Latent heat of melting of ice = 80 cal/g, i.e., the heat required to melt mg of ice = 80m cal.

so, 80m = \(\frac{mgh}{J}\)

or h = \(\frac{80 \times J}{9}=\frac{80 \times 4.2 \times 10^7}{980}\)

∴ h = 34.286 x 10^{5} cm = 34.286 cm

**Question 6. What is the temperature difference between the top and the bottom of a water fall of 100 m height if 50% of heat produced remains confined in water?**

**Answer:**

Height of the waterfall, h = 100 m = 100 x 100 cm = 1 x 10^{4} cm

Suppose, water is falling from the top to the bottom at the rate of m g/s.

Now, just before the moment it strikes the ground, its kinetic energy is equal to the loss of potential, energy = mgh.

This kinetic energy is converted fully into heat energy after striking the ground. So, heat produced = \(\frac{mgh}{J}\), where mechanical equivalent of heat, J = 4.2 x 10^{7} erg/cal.

Now, heat confined in water = \(\frac{50}{100} \times \frac{m g h}{J}=\frac{m g h}{2 J}\)

and specific heat of water, s = 1 cal • g^{-1} • °C^{-1}

If the temperature of water increases by t°C due to its fall from the top of the waterfall, then

mst = \(\frac{m g h}{2 J} \text { or, } t=\frac{g h}{2 \mathrm{sJ}}\)

∴ t = \(\frac{980 \times 1 \times 10^4}{2 \times 1 \times 4.2 \times 10^7}=0.117^{\circ} \mathrm{C}\)

**Question 7. While passing through the earth’s atmosphere, the velocity of a meteorite of mass 42 g is reduced from 10 km/s to 5 km/s. If the heat of 3.75 x 10 ^{8} cal is produced due to this change of kinetic energy, calculate the mechanical equivalent of heat.**

**Answer:**

According to the question, mass of the meteorite, m = 42kg; initial velocity, u = 10 km/s = 10 x 10^{3} m/s and final velocity, v = 5 km/s = 5 x 10^{3} m/s.

∴ loss of kinetic energy by the meteorite

= \(\frac{1}{2} m u^2-\frac{1}{2} m v^2=\frac{1}{2} m\left(u^2-v^2\right)\)

= \(\frac{1}{2} m(u+v)(u-v)\)

= \(\frac{1}{2} \times 42 \times\left(10 \times 10^3+5 \times 10^3\right) \times\left(10 \times 10^3-5 \times 10^3\right)\)

= 21 x 15 x 10^{3} x 5 x 10^{3}

= 21 X 75 x 10^{6} J

Now if J is the mechanical equivalent of heat, then the heat produced is given by

H = \(\frac{\text { loss of kinetic energy }}{J}\)

or, \(3.75 \times 10^8=\frac{21 \times 75 \times 10^6}{J}\)

∴ J = \(\frac{21 \times 75 \times 10^6}{3.75 \times 10^8}=4.2 \mathrm{~J} / \mathrm{cal}\)

**Question 8. What happens if 4g of steam at a temperature of 100 °C is passed through 100 g of water at 20°C? (Latent heat of condensation of steam = 540 cal/g.)**

**Answer:**

If 4 g of steam is transformed into water of temperature 100°C, then heat released, Q: = 540 X 4 = 2160 cal.

If temperature of 100 g water at 20°C is to be raised to 100°C, then required heat, Q2 = 100 x 1 x (100 – 20) = 8000 cal

As Q_{1} < Q_{2}, the entire amount of steam is transformed into water.

Suppose, final temperature = t°C.

According to basic principle of calorimetry, heat absorbed by water = latent heat released by steam + heat released by water at 100°C

or, 100 x 1 x (t- 20) = 540 x 4 + 4 x 1 x (100 – t)

or, 100t – 2000 = 21600 + 400 – 4t or, 100t + 4t = 2160 + 400 + 2000 or, 104t = 4560

∴ t = \(\frac{4560}{104}\) = 43.85°C

**Question 9. What amount of steam in gram is converted to water if 40 g of steam at 100°C passes through 100 g of water at 20°C? (Latent heat of condensation of steam =540 cal/g)**

**Answer:**

If 40 g of steam is transformed into water of temperature 100°C, then heat released, Q_{1} = 540 X 40 = 21600 cal

If temperature of 100 g of water is to be raised from 20°C to 100°C,then the required heat, Q_{2} = 100 x 1 x (100 – 20) = 8000 cal

As Q_{2 }< Q_{1}, entire amount of steam is not transformed into water.

Suppose, m g steam is transformed to water. From the basic principle of calorimetry, 540m = Q_{2} or, 540m = 8000

∴ m = \(\frac{8000}{540}\) = 14.81 g

**Question 10. Calculate the amount of heat required to convert 10 g of ice at -5°C into steam. (Latent heat of melting of ice = 80 cal/g, latent heat of valorisation of water = 537 cal/g and specific heat of ice = 0.5 cal • g ^{-1} • °C^{-1})**

**Answer:**

Mass of ice = 10 g.

Heat required to increase the temperature of ice from -5°C to 0°C, H_{1} = 10 x 0.5 {0-(-5)} =25 cal

Necessary heat for melting of ice, H_{2} = 10 x 80 = 800 cal

Required heat to increase the temperature of 10 g of water from 0°C to 100°C, H_{3} = 10 x 1 x (100 – 0) = 1000 cal

Heat required for vaporisation of water, H_{4} = 537 x 10 cal = 5370 cal

total heat required is given by H = H_{1} + H_{2} + H_{3} + H_{4}

= (25 + 800 + 1000 + 5370) cal = 7195 cal

**Question 11. Temperature of 200 g of water has to be raised from 24°C to 90°C by passing steam through it. Calculate the mass of steam required for it. (Latent heat of condensation of steam = 540 cal/g)**

**Answer:**

Suppose, mass of steam = m g

When m g of steam at 100°C condenses and transforms to water at 100°C, then the heat released.

H_{1} = m x 540 cal = 540m cal

Again, when temperature of m g of water at 100°C is reduced to 90°C, then heat released is given by H_{2} = m x 1 x (100 – 90) cal = 10m cal

∴ total heat released = H_{1} + H_{2} = 540m + 10m = 550m cal

Heat required to increase the temperature of 200 g of water from 24°C to 90°C, H_{3} = 200 x 1 x (90 – 40) cal = 13200 cal

According to the condition, H_{1} + H_{2} = H_{3} or, 550m = 13200

∴ m = \(\frac{13200}{550}\) = 24 g

**Question 12. Equal amounts of water and ice at 0°C temperature are mixed together, if the entire quantity of ice melts to form water at 0°C, then what is the initial temperature of water? (Latent heat of melting of ice = 80 cal/g)**

**Answer:**

Let the mass of water and ice be m g and initial temperature of water =t°C.

Latent heat required for melting of entire ice = 80m cal.

Heat released when water temperature comes down form t°C to 0°C = m x 1 x (t – 0) = m cal.

According to the basic principle of calorimetry, mt = 80m

∴ t = 80°C

**Question 13. What is the result if 2 x 10 ^{4} cal of heat is extracted for 0.4 kg of water at 30°C temperature? What is the final temperature?**

**Answer:**

Mass of water, m = 0.4 kg = 40 kg = 400 g

Specific heat of water, s = 1 cal • g^{-1} • °C^{-1}

Total heat released, Q = 2 x 10^{4} cal = 20000 cal

Now, heat released when total water at 30°C is converted to water at 0°C, Q_{1} = 400 x 1 x (30 – 0) = 12000 cal

Amount of absorbed heat, Q_{2} = Q- Q_{1} = 20000 – 12000 = 8000 cal

Now, the amount of released heat when entire amount of water of 0°C temperature is transformed to ice of 0°C is given by Q_{3} = 400 x 80 = 32000 cal

As Q_{2} < Q_{3}, total water is not transformed to ice.

Let m_{1} g water is transformed to ice.

So, m_{1} = Q_{2} or, 80m_{1} = 8000

∴ m_{1} = 100 g

Hence, if 2 x 10^{4} cal of heat is extracted from 0.4 kg of water at 30°C temperature, final temperature is 0°C, in which 100 g of ice and (400 – 100) g = 300 g of water is available.

**Question 14. How do you divide 1 kg of water at 5°C temperature in two parts, such that if one part is transformed into ice, then the released heat is sufficient to transform the other part into vapour? (Latent heat of melting of ice = 80 cal/g and latent heat of vaporisation of water = 540 cal/g .)**

**Answer:**

Mass of water = 1 kg = 1000 g

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

Suppose, Mg of water becomes ice and (1000 -M) g of water is vaporised.

Now when M g of water at 5°C is transformed to ice at 0°C, then the heat released is given by

Q_{1} = M x 1 x (5 – 0) + 80M = 5M + 80M = 85Mcal

Again, heat required to transform (1000 – M) g of water at 5°C to steam of 100°C,

Q_{1} = (100 – M) x 1 x (100 – 5) + 540 x (1000 – M) = 635 x (1000 -M) cal

According to the condition, Q_{1} = Q_{2}

or, 85 M = 635 X (1000 – M)

or, 85M = (635 x 1000) – 635M

or, SSM + 635M= 635 X 1000

or, 720M = 635 x 1000

∴ M = 881.94 g

and 1000 – M = 1000 – 881.94 = 118.06 g

Hence, the amount of heat that is released when 881.94 g of water is transformed into ice is sufficient to transform 118.06 g of water into vapor.

**Question 15. 168.75g of water is produced when steam at 100°C is passed through a chunk of ice at 0°C. The mass of ice before the passing of steam is 450 g and after the passing of steam the mass becomes 300 g. What is the latent heat of the vaporization of steam? [Latent heat of melting of ice = 80 cal/g]**

**Answer:**

The mass of ice before the passing of steam is 450 g and after the passing of steam the mass becomes 300 g.

Therefore, mass of melted ice = (450 – 300) = 150 g

So, heat required for melting of 150 g of ice, Q_{1} = 150 x 80 = 12000 cal

Mass of condensed steam = (168.75- 150) g = 18.75 g

Let the latent heat of vaporization of steam = L cal/g.

Now when 18.75 g of steam at 100°C is converted to water at 0°C, then the heat released is given by

Q_{2} = 18.75 L + 18.75 x 1 x (100 – 0)

= 18.75 L+ 1875 cal

According to the question,

Q_{1} = Q_{2} or, 12000 = 18.75L + 1875

or, 18.75L = 12000 – 1875

or, 18.75L = 10125

∴ L = \(\frac{10125}{18.75}\) = 540 cal/g

**Question 16. What amount of heat is required to convert 10 g of ice at 0°C to 10 g of water at 10°C?**

**Answer:**

The total amount of heat is required = heat is required for melting of ice + heat is required to increase temperature from 0°C to 10°C

= m • L + m • s • (t_{2} – t_{1})

= 10 x 80 + 10 x 1 x (10 – 0)

= 800 + 100 = 900 cal

**Question 17. What amount of heat is required to convert 10 g of water at 0°C to water vapor at 100°C.**

**Answer:**

The total amount of heat is required = heat is required to increase the temperature from 0°C to 100°C + heat is required for vaporization

= m • s • (t_{2} – t_{1}) + m • L

= 10 • 1 • (100 – 0) + 10 X 537 = 1000 + 5370 = 6370 cal