Heat Solutions Class 10

BEGUN PUBLISHING —-

The following heat solutions class 10 guide has questions from the Science Class 10 Textbook published by the Ministry of Education, Government of Nepal. Before looking onto this solution, we recommend you to have a look at Heat Notes.

A. Choose the best alternatives.

1. How many joules make one calorie?

1. 8.2 J
2. 4.2 J
3. 1000 J
4. 4200 J

Reason: We know, 1 calorie (cal) = 4.2 joules

2. What are the boiling point of mercury and alcohol?

1. 357^oC and 78^oC
2. -39^oC and -115^oC
3. 100^oC and 120^oC
4. 90^oF and 100^oF

Reason: Boiling point of mercury and alcohol is 357^oC and 78^oC. Similarly, their melting point is -39^oC and -117^oC.

3. What is the sum of kinetic energy of the molecules of a system called?

1. Heat
2. Temperature
3. Specific heat capacity
4. Heat equation

Reason: Heat is the sum of kinetic energy of the molecules of a system. And, temperature is the average kinetic energy of the molecules of a system.

4. Which of the following is the specific heat capacity of water?

1. 200 J/kg^oC
2. 4000 J/kg^oC
3. 4200 J/kg^oC
4. 820 J/kg^oC

Reason: In SI system, specific heat capacity of water is 4200 J/kg^oC and that in CGS system is 1 cal/gm^oC.

5. What is the melting point of water in Fahrenheit scale?

1. 100 ^oF
2. 0 ^oF
3. 373 ^oF
4. 32 ^oF

Reason: Melting point of water in Celsius scale is 0^oC. And, 0^oC is equal to 32^oF. Or, melting point of ice in Fahrenheit is 32^oF and boiling point of water is 212^oF.

6. What is our normal body temperature?

1. 98.6 ^oF
2. 35 ^oC
3. 100 ^oC

Reason: Normal human body temperature is 37^oC or 98.6^oF.

B. Answer the following questions.

1. Define heat on the basis of kinetic theory of molecules. List out the differences between heat and temperature.

Answer: On the basis of kinetic theory of molecules, heat is defined as the sum of kinetic energy of all the molecules present in a given body.

Following are the major differences between heat and temperature:

Heat	Temperature
Heat is the sum of kinetic energy of all the molecules contained in a body.	Temperature is the average kinetic energy of the molecules in a body.
It is the total energy in flow between two bodies.	It is the measure of how hot or cold an object is.
It is a cause.	It is an effect.
SI unit of heat is Joule (J).	SI unit of temperature is Kelvin (K).
It is a derived physical quantity.	It is a fundamental physical quantity.

2. On which factor does quantity of heat depend? What is the effect on the molecules of matter when the temperature is varied?

Answer: Quantity of heat of a body depends upon the following two factors:

1. Number of molecules present in the body
2. Average kinetic energy of the molecules

In general, when the temperature is varied, there is change in kinetic energy of the molecules of the body. On increasing the temperature, the molecules start vibrating about their mean position and their average kinetic energy increases. In comparison, the average kinetic energy of molecules of a hot body is found to be more than that of a cold body.

3. What is thermometric liquid? On which principle is a thermometer based?

Answer: Any liquid that is used in a thermometer as a thermometric substance is called thermometric liquid.

A thermometer is based on the principle that “body expands on heating and contracts on cooling”. When the mercury used in mercury thermometer gains heat, it increases its volume. And, due to the organized calibration of the thermometer, the temperature of the measured body can be read easily.

4. Write down the advantages and disadvantages of the mercury thermometer.

Answer: Following are the advantages of mercury thermometer:

1. It can measure a wide range of temperature ranging from -39^oC to 357^oC.
2. Mercury expands uniformly over wide range of temperature.
3. The specific heat capacity of mercury is low.
4. Mercury has a distinct colour, i.e., silvery opaque.

Following are the disadvantages of mercury thermometer:

1. Mercury is a toxic liquid so it should be handled with care.
2. It cannot be used to measure the temperature of very cold regions.

___ COMPLETED ___

5. Describe the structure and function of the maximum-minimum thermometer.

Answer: Maximum-minimum thermometer has a U-shaped tube which contains mercury column and alcohol is put over the mercury column.

It is used especially by the meteorologists to measure the maximum and minimum temperature of a place.

6. Define specific heat capacity. What are specific heat capacities of water and copper?

Answer: Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of unit mass of that substance through 1Celsius degree.

Specific heat capacity of water is 4200 J/kg^oC and that of copper is 400 J/kg^oC.

7. When equal mass of hot water and oil at the same temperature are allowed to cool, water cools slower than the oil. Why?

Answer: When both hot water and oil have equal masses, due to higher specific heat capacity of water (4200 J/kg^oC) than that of oil (2000 J/kg^oC), it requires to lose higher quantity of heat to bring 1 C^o change in its temperature as compared to the oil. As a result, when the same amount of heat has been taken out, the change in temperature of water is less than that of the oil.

8. What is heat equation? On which four factors does the quantity of heat lost or gained by a body depend?

Answer: Heat equation is a mathematical relation between quantity of heat gained or lost by a body (Q), difference in temperature (dt), mass (m) of the body, and specific heat capacity (s) of the body for a definite change in its temperature.

Q = ms ‘dt’

Following are the four factors on which the quantity of heat lost or gained by a body depend:

1. Mass of the body (m)
2. Specific heat capacity of the body (s)
3. Initial temperature (θ₁)
4. Final temperature (θ₂)

9. Study the following table and answer the questions, below.

Substances	Specific heat capacity (J/kgoC)
A	1000
B	450
C	780

• If equal amount of heat is supplied to equal masses of all these substances at same temperature, which one will have the lowest temperature? Why?
• Which one of them is suitable for cooling and heating purpose? Why?
• Which one will penetrate to the maximum depth when equal mass of these substances are heated to make their final temperature 100^oC and then kept on a wax slab of uniform thickness? Why?
• What do you mean by specific heat capacity of B is 450 J/kg^oC?

Answer: According to the above table,

• At same initial temperature, equal amount of heat and equal mass, rise is temperature of bodies is inversely proportional to its specific heat capacity. Thus, the body with high specific capacity, i.e., substance A will have the lowest temperature in the given condition.
• A body that gains heat easily and that can cool down easily is preferred for cooling and heating purposes. For this, the body B whose specific heat capacity is the lowest is preferred.
• When the bodies are heated to a final temperature 100^oC having equal mass, then the amount of heat contained by them is directly proportional to their specific heat capacity. Since body with higher specific heat capacity has the ability to penetrate deeper into the wax slab, body A will do it more.
• The specific heat capacity of B is 450 J/kg^oC. This means that to increase the temperature of the substance B through 1 ^oC, we need to supply 450 joules of heat for every 1 kilogram mass of the substance.

C. Solve the following numerical problems.

1. What amount of heat is required to raise the temperature of 5 kg of water through 20^oC?

Solution:

Here,

Mass of water (m) = 5 kg

Difference in temperature (dt) = 20^oC

Specific heat capacity of water (s) = 4200 J/kg^oC

Now,

Amount of heat required (Q) = ms ‘dt’

= 5 x 4200 x 20

= 420 000

= 4.2×10⁵ J

Hence, the amount of heat required in the above condition is 4.2×10⁵ J.

2. How much heat is required to raise the temperature of 160 gm of an iron ball from 25^oC to 15^oC? (Specific heat capacity of iron is 480 J/kg^oC)

Solution:

Here,

Mass of iron ball (m) = 160 gm = 0.16 kg

Initial temperature (θ₁) = 25^oC

Final temperature (θ₂) = 15^oC

Specific heat capacity of iron (s) = 480 J/kg^oC

Now,

Since, we are decreasing the temperature of iron ball, we need to absorb heat from the ball.

Q = ms ‘dt’

= ms ‘θ₂ – θ₁’

= 0.16 * 480 * (15 – 25)

= -768 J

The minus sign in the answer tells us that we need to absorb 768 Joules of heat from the ball.

3. When 456 KJ of heat is supplied to a copper ball of mass 2 kg at 30^oC, its final temperature is found to be 600^oC. Calculate its specific heat capacity.

Solution:

Here,

Amount of heat (Q) = 456 KJ = 456 000 J

Mass of ball (m) = 2 kg

Initial temperature (θ₁) = 30^oC

Final temperature (θ₂) = 600^oC

Change in temperature = (θ₂ – θ₁) = 600 – 30 = 570 ^oC

Now,

Q = ms ‘dt’

or, 456 000 = 2 x s x 570

or, 456 000 = 1140 s

or, s = 456 000 / 1140

Therefore s = 400 J/kg^oC

Hence, the required specific heat capacity of copper is 400 J/kg^oC.

4. When 2 kg of water is heated by an electric heater of 1 KW for some time, it supplied 84 KJ of heat. If the specific heat capacity of water is 4200 J/kg^oC, calculate its change in temperature.

Solution:

Given,

Mass of water (m) = 2 kg

Power of heater (P) = 1 KW = 1000 W

Amount of heat (Q) = 84 KJ = 84 000 J

Specific heat capacity of water (s) = 4200 J/kg^oC

To find: change in temperature (dt) = ?

We know,

Q = ms ‘dt’

or, 84 000 = 2 x 4200 x dt

or, 84 000 = 8 400 x dt

or, dt = 84 000 / 8 400

Therefore dt = 10^oC

Hence, the required change in temperature of water is 10^oC.

5. If an electric heater of 3 KW used to heat 100 kg of water at 30^oC and it supplied 3.6 x 10⁶J of heat energy, calculate the final temperature of water.

Solution:

Given,

Mass of water (m) = 100 kg

Initial temperature of water (t₁) = 30^oC

Amount of energy (Q) = 3.6 x 10⁶ J

Specific heat capacity of water (s) = 4200 J/kg^oC

To find: final temperature of water (t₂) = ?

We know,

Q = ms ‘dt’  where dt = t₂ – t₁

or, 3.6 x 10⁶ = 100 x 4200 x (t₂ – 30)

or, 3 600 000 = 420 000 (t₂ –30)

or, 3 600 000 / 420 000 = t₂ – 30

or, 8.57 = t₂ – 30

or, t₂ = 30 + 8.57

Therefore, 1₂ = 38.57 ^oC

Hence, the required final temperature of water is 38.57^oC.

6. 80 KJ of heat is supplied to an object of mass 4 kg at 30^oC. If the specific heat capacity is 1000 J/kg^oC, what will be the final temperature?

Solution:

Here,

Amount of heat (Q) = 80 KJ = 80 000 J

Mass of object (m) = 4 kg

Initial temperature of object (t1) = 30 ^oC

Specific heat capacity (s) = 1000 J/kg^oC

To find: final temperature (t2) = ?

We know,

Q = ms (t2-t1)

or, 80 000 = 4 x 1000 x (t2 – 30)

or, 80 000 / 4000 = t2 – 30

or, 20 = t2 – 30

or, t2 = 20 + 30

Therefore t2 = 50^oC

Hence, the required final temperature of the object is 50^oC.

7. 10 kg of water at 90^oC is cooled by mixing 20kg of water at 20^oC. What is the final temperature of the mixture?

Solution:

Here,

Mass of water (m1) = 10 kg

Initial temperature of water (t1) = 90^oC

Mass of other water (m2) = 20 kg

Initial temperature of other water (t2) = 20^oC

Let t be the final temperature of the mixture (t) = ?

Since t1 > t2, heat flows from water (m1) to the other water (m2). So, t1 > t > t2

From principle of calorimetry,

Amount of heat lost = Amount of heat gained

or, m1 s (t1 – t) = m2 s (t – t2)

Both bodies are water so their specific heat capacity is the same and we can cancel them.

or, m1 (t1 – t) = m2 (t – t2)

or, 10 (90 – t) = 20 (t – 20)

or, 90 – t = 20 / 10 x (t – 20)

or, 90 – t = 2 (t – 20)

or, 90 – t = 2t – 40

or, 90 + 40 = 2t + t

or, 130 = 3t

Therefore, t = 43.33^oC

Hence, the required final temperature of the mixture is 43.33^oC.