Physics questions, chapter 4

Nama / NameKelas / ClassTarikh / Date

Heat

46 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first.

Section A: objective questions

10 marks

Choose the best answer for each question.

Question 1

Which statement describes two objects in thermal equilibrium?

One object has a higher heat capacity than the other
Heat flows from the colder object to the hotter object
The net rate of heat transfer between them is zero
Both objects have the same mass

Show answerHide answer

At thermal equilibrium the two objects are at the same temperature. Heat flows both ways at the same rate, so the net heat transfer is zero (textbook p. 121).

For the teacher: Remembering, low difficulty, standard 4.1

Textbook p. 121

Question 2

Which pair of fixed points is used to calibrate a liquid-in-glass thermometer?

Melting point of ice and boiling point of pure water
Boiling point of water and room temperature
Body temperature and melting point of ice
Freezing point of water and body temperature

Show answerHide answer

The two fixed points are the melting point of ice (0 °C) and the boiling point of distilled water (100 °C).

For the teacher: Remembering, low difficulty, standard 4.1

Textbook p. 123

Question 3

An uncalibrated thermometer has a length of 20 cm between the ice point (0 °C) and the steam point (100 °C). When the bulb is put in a liquid, the liquid column is 6 cm long, measured from the ice point. What is the temperature of the liquid?

6 °C
30 °C
60 °C
70 °C

Show answerHide answer

θ = (6 ÷ 20) × 100 = 30 °C.

For the teacher: Applying, low difficulty, standard 4.1

Textbook p. 124

Question 4

Water is used as the cooling agent in a car radiator because water

has a low boiling point
has a low density
is a good conductor of electricity
has a high specific heat capacity

Show answerHide answer

Water has a high specific heat capacity, so it absorbs a large amount of heat from the engine with only a small rise in temperature.

For the teacher: Understanding, low difficulty, standard 4.2

Textbook p. 132

Question 5

Equal masses of water, aluminium, iron and copper receive the same amount of heat. Which one has the greatest rise in temperature? (Specific heat capacities: water 4 200, aluminium 900, iron 450, copper 390 J kg⁻¹ °C⁻¹)

Copper
Iron
Aluminium
Water

Show answerHide answer

Δθ = Q ÷ mc. For the same Q and m, the substance with the smallest c has the largest Δθ, which is copper.

For the teacher: Applying, medium difficulty, standard 4.2

Textbook p. 128

Question 6

A liquid boils at a constant temperature although heat is still supplied. What happens to the heat absorbed?

It increases the average kinetic energy of the molecules
It overcomes the forces of attraction between the molecules
It increases the mass of the liquid
It lowers the boiling point of the liquid

Show answerHide answer

The latent heat is used to break the bonds and separate the molecules, not to raise their average kinetic energy. So the temperature stays constant.

For the teacher: Understanding, medium difficulty, standard 4.3

Textbook p. 138

Question 7

The air in a sealed syringe has a volume of 30 cm³ and a pressure of 200 kPa. The temperature stays constant. What is the pressure when the volume is reduced to 15 cm³?

100 kPa
215 kPa
400 kPa
6 000 kPa

Show answerHide answer

Boyle's law: P₁V₁ = P₂V₂, so P₂ = (200 × 30) ÷ 15 = 400 kPa.

For the teacher: Applying, low difficulty, standard 4.4

Textbook p. 151

Question 8

A fixed mass of gas has a volume of 250 cm³ at 27 °C. It is heated to 87 °C at constant pressure. What is the new volume?

300 cm³
310 cm³
805 cm³
1 000 cm³

Show answerHide answer

Convert to kelvin: T₁ = 300 K, T₂ = 360 K. V₂ = 250 × 360 ÷ 300 = 300 cm³. Using °C directly would give a wrong value of about 805 cm³.

For the teacher: Applying, medium difficulty, standard 4.4

Textbook p. 156

Question 9

A gas in a rigid container is at 27 °C. At what temperature will its pressure be twice the original value?

54 °C
327 °C
600 °C
300 °C

Show answerHide answer

At constant volume P ∝ T (kelvin). T₁ = 300 K, so T₂ = 600 K = 327 °C. Doubling the Celsius value (54 °C) is a common error.

For the teacher: Analysing, high difficulty, standard 4.4

Textbook p. 160

Question 10

Diagram 1 shows the graph of pressure against temperature for a fixed mass of gas at constant volume. At what temperature does the straight line, extended, meet the axis where P = 0?

Graph of pressure against temperature in degrees Celsius; points at 30, 50, 70 and 90 °C lie on a straight line that rises from about 0 kPa at the far left — Diagram 1

−373 °C
−100 °C
0 °C
−273 °C

Show answerHide answer

Extrapolating the graph to P = 0 gives absolute zero, −273 °C (0 K).

For the teacher: Analysing, low difficulty, standard 4.4

Textbook p. 155

Section B: structured questions

36 marks

Answer all questions. Write your answers in the spaces provided.

Question 11

A clinical thermometer is placed under the tongue of a patient.

[1 mark] (a)What is meant by thermal equilibrium?

Show answerHide answer

Two objects in thermal contact have the same temperature and the net heat transfer between them is zero1

Tick each point you got.

[2 marks] (b)Explain why the thermometer must be left under the tongue for a short time before it is read.

Show answerHide answer

Heat flows from the body to the thermometer1
until both have the same temperature (thermal equilibrium), so the reading is the body temperature1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 4.1

Textbook p. 122

Question 12

0.30 kg of water at 80 °C is mixed with 0.60 kg of water at 20 °C in an insulated container. [Specific heat capacity of water = 4 200 J kg⁻¹ °C⁻¹]

[1 mark] (a)State the principle used to find the final temperature.

Show answerHide answer

Heat released by the hot water = heat absorbed by the cold water1

Tick each point you got.

[3 marks] (b)Calculate the final temperature of the mixture.

Show answerHide answer

0.30 × 4 200 × (80 − y) = 0.60 × 4 200 × (y − 20)1
24 − 0.3y = 0.6y − 12 (or equivalent)1
y = 40 °C1

Tick each point you got.

[1 mark] (c)State one assumption you made.

Show answerHide answer

No heat is lost to (or gained from) the surroundings1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 4.2

Textbook p. 135

Question 13

Explain each of the following using specific heat capacity.

[2 marks] (a)The body of a cooking pot is made of aluminium but its handle is made of plastic.

Show answerHide answer

Aluminium has a low specific heat capacity so the pot heats up quickly1
Plastic has a high specific heat capacity so the handle heats up slowly and is safe to hold1

Tick each point you got.

[2 marks] (b)A sea breeze blows from the sea to the land during the day.

Show answerHide answer

Land has a lower specific heat capacity than the sea, so land gets hotter faster; air above it is hot and rises1
Cooler air from the sea moves in to take its place1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 4.2

Textbook p. 134

Question 14

A solid of mass 0.20 kg is heated by a 50 W heater. Diagram 2 shows how its temperature changes with time. Assume all the heat from the heater is absorbed by the substance.

Graph of temperature against time: 50 °C at 0 min rising to 80 °C at 3 min, flat at 80 °C until 9 min, then rising to 104 °C at 12 min — Diagram 2

[1 mark] (a)What is the melting point of the substance?

Show answerHide answer

80 °C1

Tick each point you got.

[2 marks] (b)Explain why the temperature stays constant from Q to R although heat is still supplied.

Show answerHide answer

Heat is absorbed as latent heat to weaken the bonds between molecules1
It does not increase the average kinetic energy of the molecules1

Tick each point you got.

[2 marks] (c)Calculate the heat absorbed from Q to R.

Show answerHide answer

Time = (9 − 3) × 60 = 360 s1
Q = Pt = 50 × 360 = 18 000 J1

Tick each point you got.

[2 marks] (d)Calculate the specific latent heat of fusion of the substance.

Show answerHide answer

l = Q ÷ m = 18 000 ÷ 0.201
= 9.0 × 10⁴ J kg⁻¹1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 4.3

Textbook p. 138

Question 15

An electric kettle of power 1.5 kW changes 0.40 kg of water at 25 °C into steam at 100 °C. [c of water = 4 200 J kg⁻¹ °C⁻¹; l_v of water = 2.26 × 10⁶ J kg⁻¹]

[2 marks] (a)Calculate the heat needed to raise the temperature of the water to 100 °C.

Show answerHide answer

Q₁ = mcΔθ = 0.40 × 4 200 × 751
= 126 000 J1

Tick each point you got.

[2 marks] (b)Calculate the heat needed to change the water at 100 °C into steam.

Show answerHide answer

Q₂ = ml_v = 0.40 × 2.26 × 10⁶1
= 9.04 × 10⁵ J1

Tick each point you got.

[3 marks] (c)Calculate the total heat and the time taken by the kettle.

Show answerHide answer

Total Q = 126 000 + 904 000 = 1.03 × 10⁶ J1
t = Q ÷ P = 1 030 000 ÷ 1 5001
= 687 s (about 11.4 minutes)1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 4.3

Textbook p. 147

Question 16

The end of a syringe is closed. It contains 60 cm³ of air at a pressure of 102 kPa. The piston is pushed in slowly until the volume is 24 cm³.

[1 mark] (a)State Boyle's law.

Show answerHide answer

The pressure of a fixed mass of gas is inversely proportional to its volume at constant temperature1

Tick each point you got.

[2 marks] (b)Calculate the pressure of the compressed air.

Show answerHide answer

P₁V₁ = P₂V₂: 102 × 60 = P₂ × 241
P₂ = 255 kPa1

Tick each point you got.

[2 marks] (c)Explain, using the kinetic theory of gas, why the pressure increases.

Show answerHide answer

The same number of molecules is in a smaller volume, so there are more molecules per unit volume1
Molecules collide with the walls more frequently, so the force per unit area increases1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 4.4

Textbook p. 151

Question 17

Use the gas laws to answer the following.

[1 mark] (a)Convert 17 °C to kelvin.

Show answerHide answer

17 + 273 = 290 K1

Tick each point you got.

[2 marks] (b)An air bubble under a leaf has a volume of 3.0 cm³ at 17 °C. Calculate its volume when the water warms to 37 °C. Pressure is constant.

Show answerHide answer

V₁/T₁ = V₂/T₂: 3.0 ÷ 290 = V₂ ÷ 3101
V₂ = 3.21 cm³1

Tick each point you got.

[2 marks] (c)A sealed gas cylinder at 27 °C has a pressure of 250 kPa. It is heated to 177 °C. Calculate the new pressure.

Show answerHide answer

P₁/T₁ = P₂/T₂: 250 ÷ 300 = P₂ ÷ 4501
P₂ = 375 kPa1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 4.4

Textbook p. 163

Back to notes Chapter 5 questions Model papers