full solution of ch 02

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NSS Physics in Life Full Solution of Textbooks(Heat and Gases)

Chapter 2 Heat and Internal Energy

Checkpoint (p.36)

1. C

2. (a) Incorrect. The internal energy of a body is related to its temperature, mass,

material and state. Thus, a body of a higher internal energy is not

necessarily the one of a higher temperature.

(b) Incorrect. The internal energy is also related to the material of a body.

Although the two cups of liquid have the same mass, temperature and state,

they have different internal energy.

(c) Correct

(d) Incorrect. Temperature is a measure of the average kinetic energy due to

random motion of the molecules of a body. Since the two liquids are at the

same temperature, the average kinetic energies due to random motion of the

molecules of the two liquids are the same.

3. (a) (1)

(b) (2)

(c) (1)

4. Peter is incorrect.

Heat always flows from a body of a higher temperature to another body of a

lower temperature. Heat is the energy transferred due to the temperature

difference between two bodies while internal energy is the energy stored in a

body.

Checkpoint (p.39)

1. B

2.

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3. (a) Energy consumed by two air conditioners

hkW 12

h 4kW 5.12

(b) Energy consumed by the tree lights

(c) Energy consumed by the washing machine

Therefore, in cases (a) and (c), 12 kW h of energy is consumed by the electrical

appliance.

Exercise (p.39)

1. B

Internal energy of a body is the sum of the kinetic energy due to random motion

and the potential energy of all its molecules. It is not the sum of the kinetic

energy and potential of the body.

2. A

3. A

4. (a) (1) The molecular average kinetic energy of the milk increases when the

temperature of the milk increases.

(2) The molecular potential energy of the milk does not change because

the state of the milk does not change in the process. (Assume the

thermal expansion of the milk is negligible.)

(3) The internal energy of the milk increases when the temperature of the

milk increases.

(b) (1) The molecular average kinetic energy of the water does not increase

because the temperature of the water remains unchanged.

(2) The molecular potential energy of the water increases when the water

changes from the liquid state to the gaseous state.

(3) The internal energy of the water increases when water changes from

the liquid state to the gaseous state.

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(c) (1) Heating a gas inside a rigid container will increase its temperature.

Thus, the molecular average kinetic energy of the gas increases in the

process.

(2) Since the container is rigid, the average distance between the gas

molecules remains unchanged in the process. Thus, the molecular

potential energy of the gas does not increase.

(3) The internal energy of the gas increases when the temperature of the

gas increases.

5. (a) Temperature is a physical quantity that measures the degree of hotness or

coldness of an object. Internal energy is the energy stored in a body.

Microscopically, temperature is a measure of the average kinetic energy due

to random motion of the molecules of a body. The internal energy of a body

is the sum of the kinetic energy due to random motion and the potential

energy of its molecules.

(b) I agree with Jane.

Internal energy is the energy stored in a body. It is related to the mass,

material, temperature and state of the body. It increases with the temperature

of the body. However, an increase in internal energy does not necessarily

imply an increase in temperature.

6. (a) Heat is the energy transferred from one body to another as a result of the

temperature difference between the bodies.

Internal energy is the energy stored in a body.

(b) His statement is incorrect.

Heat is the energy transferred from one body to another due to the

temperature difference between the two bodies. Heat flows from a body of

higher temperature to another body of lower temperature when the two

bodies are brought in contact.

7. (a) Applying ,

The power rating of the heater is 1500 W.

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(b) ,Applyingt

EP

It takes 833 s to transfer 1 MJ of energy to the liquid.

Checkpoint (p.41)

1. (c), (a), (b), (d)

2. Applying , the energy removed is

Thus, an amount of 24 000 J of energy should be removed from the can of cola.

Checkpoint (p.46)

1. A

2. (a) W

(b) Z

Exercise (p.46)

1. A

2. (a) Applying ,

The heat capacity of the oil is 1940 J °C−1.

(b) Applying ,

The specific heat capacity of the oil is 2420 J kg−1 °C−1.

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3. (a) Applying ,

The energy loss to the surroundings is 29 400 J.

(b) Applying ,

The average rate of energy lost by the soup is 17.5 W.

4. (a) Let E be the energy provided by the whole pack of oat flakes.

By proportion,

The energy provided by the whole package is 1.06 × 107 J.

(b) Applying ,

Using 1.06 × 107 J of energy can heat 38.74 kg of water from 25°C to 90°C.

5. Applying , the energy gained by the sushi when its temperature is

raised from 2°C to 22°C is

Applying ,

Jack must eat the sushi within 71.4 minutes after taking it out of the refrigerator.

6. (a) Applying , energy removed from the air is

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Applying ,

The air conditioner is switched on for 14.8 minutes.

(b) The energy drawn out

Applying ,

The temperature rise of the pool of the water will be 9.84°C.

7. (a) The value of the specific heat capacity of water obtained will be larger than

the standard value. This is because there is heat lost to the surroundings.

(b) (1) Covering the cup with a lid can achieve a more accurate result in the

experiment. This is because heat lost to the surrounds is reduced.

(2) Using more water cannot achieve a more accurate result in the

experiment. When more water is used, a smaller change in the water

temperature is obtained. The percentage error in measuring the

temperature is larger and hence affects the accuracy of the results.

(3) Immersing the heater completely in the water can achieve a more

accurate result in the experiment. This is because most of the energy

supplied from the heater is transferred to the water, instead of the air.

8. (a) Applying ,

The specific heat capacity of the metal is 809 J kg−1 °C−1.

(b) The value obtained in (a) is much higher than the actual specific heat

capacity of the metal. This is because part of the energy supplied by the

heater is transferred to the surroundings.

(c) (Any reasonable answer)

(1) Add a few drops of oil to the holes in the metal block to enhance heat

transfer between the block, the heater and the thermometer.

(2) Place a wooden block beneath the metal block to reduce heat lost to the

bench through conduction.

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Checkpoint (p.50)

1. C

2. Ivan’s statement is correct while Joe’s statement is incorrect.

The law of conservation of energy only states that when two bodies are brought

in contact, the energy lost by a hotter body is equal to the energy gained by the

colder body if there is no heat lost to the surroundings.

The forms of energy involved in the process depend on individual situation.

Checkpoint (p.52)

1. D

2. (a) incorrect

(b) incorrect

(c) correct

Exercise (p.53)

1. B

2. (a) Let m be the mass of the milk added to the hot tea.

Energy Q1 gained by the milk

Energy Q2 lost by the tea

By the law of conservation of energy, we have

Therefore, 0.0183 kg of milk has been added.

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(b) Let T be the final temperature of the tea.

Energy gained by the two slices of lemon

Energy lost by the tea


Thus, the total energy transferred from the tea to the two slices of lemon

3. (a) Since the density of water is 1 kg L1, the mass of the hot water and tap

water flowing out of the tap per second is 0.05 kg and 0.15 kg respectively.

Let T be the final temperature of the mixture.

Energy Q1 gained by the tap water

Energy Q2 lost by the hot water


The final temperature of the mixture is 36.25°C.

(b) Let M be the mass of the hot water added.

Energy Q1 gained by the tap water

Energy Q2 lost by the hot water

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By the conservation of energy, we have

He should add 12 kg of hot water for the final water temperature to be 40°C.

(c) The final water temperature is lower than 40°C. This is because part of the

energy is lost to the surroundings.

4. (a) Since the metal is heated by boiling water for some time, its initial

temperature should be 100°C. The metal block is then transferred to the

water quickly.

Let c be the specific heat capacity of the metal.

Energy lost by the metal

Since the density of water is 1 kg L1, the mass of 250 mL water is 0.25 kg.

Energy gained by the water

By the conservation of energy, we have

The specific heat capacity of the metal is 360 J kg−1 °C−1.

(b) (i) It is a possible reason.

If the metal is still hotter than the water when the thermometer reading

is taken, it means heat is still transferring from the metal to the water.

The energy transferred to the water is underestimated. Thus, the

measured value of the specific heat capacity of the metal is smaller

than the actual value.

(ii) It is not a possible reason.

This is because the boiling water adhered to the block will transfer heat

to the water in the cup. More amount of energy is transferred to the

water. Thus, the measured value of the specific heat capacity of the

metal should be larger than the actual value.

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(iii) It is a possible reason.

The metal block loses heat to the surroundings when it is transferred.

The energy gained by the water is smaller than the energy lost by the

metal block. Thus, the measured value of the specific heat capacity of

the metal is smaller than the actual value.

5. hot water pack

Chapter Exercise (p.56)

1. C

2. A

3. B

4. B

5. A

6. C

7. A

8. (a) Let T be the final temperature of the glass envelope.

Energy gained by the glass envelope

(1A)

According to Table 2.1 on p.41, the specific heat capacity of glass is

840 J kg−1 °C−1. Applying , we have

(1M + 1A)

The final temperature of the glass envelope is 34.5°C.

(b) When the temperature of the glass envelope rises, the temperature

difference between the glass envelope and the surroundings will become

larger. Thus, the rate of energy lost to the surroundings will increase. (1A)

When the rate of energy gained by the glass envelope is equal to the rate of

energy lost to the surroundings, the temperature of the glass envelope will

stop rising. (1A)

9. Applying , the energy gained by the air per second is

(1M + 1A)

Part of electrical energy consumed is transformed into the kinetic energy of the

moving air and the internal energy of the hair dryer and part of the energy is lost

to the surroundings through radiation. Therefore, the electrical energy consumed

is different from the energy gained by the air. (2A)

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10. The mass of air in the room

Applying , the energy required to raise the air temperature to 20°C is

(1M + 1A)

Let t be the time taken for the air temperature to rise to 20°C.

Applying , we have

(1M)

It takes 0.336 h for the air temperature to rise to 20°C. (1A)

11. Applying , the energy lost by the soup in the bowl is

(1A)

Let T be the final temperature of soup in the bowl.

Applying , we have

(1A)

Therefore, the temperature of the soup in the bowl is 65.7°C after 45 minutes.

Let T1 be the temperature of the soup after the soup is poured into the cup and

T2 be the final temperature of the soup in the cup after 45 minutes.


(1A)

Applying , the energy lost by the soup in the cup is

Applying , we have

(1A)

The temperature of the soup in the cup is 80.0°C after 45 minutes.

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12. (a) The temperature of the aluminium block increases almost linearly from

20°C to 27°C in the first 3 minutes.

Let c be the specific heat capacity of aluminium.

Assume that there is no energy lost to the surroundings. The energy gained by the aluminium block is (1M)

Applying , we have

(1A)

From the first 3 minutes of the graph, the specific heat capacity of

aluminium is 983 J kg−1 °C−1.

(b) Let E be the energy lost to the surroundings in 8 minutes.

According to the law of conservation of energy and applying , we

have

(1M + 1A)

The energy lost to the surroundings in 8 minutes is 7536 J.

(c) As the temperature of the aluminium block increases, the rate of heat lost to

the surroundings gradually increases (1A). The effective power in heating the

aluminium decreases (1A) and the slope of temperature-time graph decreases

accordingly.

13. (a) From Table 2.1 on p.41, the specific heat capacity of copper is

390 J kg−1 °C−1.

Applying , the energy lost by the block is

(1M + 1A)

(b) Since the density of water is 1 kg L−1, the mass of 250 mL water is 0.25 kg.

Applying , the energy gained by the water is

(1M + 1A)

(c) The percentage of energy lost by the copper block to the surroundings is

(1M + 1A)

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14. (a) Since the density of water is 1 kg L−1, the mass of 250 mL water is 0.25 kg.

Let T be the final temperature of the mixture.

Applying , the energy Q1 lost by the water is

Applying , the energy Q2 gained by the glass and the egg is


(1M + 1A)

The final temperature of the mixture is 81.1°C which is lower than the

suggested temperature 84°C. It is not hygienic to take in the drink. (1A)

(b) (Any reasonable answers)

(1) Using more water at 96°C. (1A)

(2) The glass is preheated before the hot water is poured into it. (1A)

15. (OCR Higher 2003 Specimen)

16. (HKCEE 2000 P1 Q8)

17. (a) A dry roof will heat up more quickly than a wet roof in daytime. This is

because a wet roof has more water than a dry roof (1A).

In daytime, a roof will absorb heat from the sunshine. Since water has a

high specific heat capacity, it absorbs a large amount of energy with only a

small temperature change (1A). Thus, the wet roof has a lower temperature

change than the dry roof in daytime.

(b) This method can also be used at night (1A). At night, a roof will release heat

to the surroundings (1A). Since water has a high specific heat capacity, it

releases a large amount of energy with only a small temperature change.

Thus, the wet roof will cool down more slowly than the dry roof at night

(1A).

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full solution of ch 02

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