Lesson 9.1.1
9.1.1 Specific heat capacity and specific latent heat Quiz: Pearson Edexcel Physics, Unit 9
20 questions
In partnership with Revision Ninja
Lesson 9.1.1, Specific heat capacity and specific latent heat: 20 multiple choice questions for the Pearson Edexcel Physics (9PH0), Unit 9: Thermodynamics, written with Revision Ninja.
Host it live on the board and students join with a game code on their own devices, or revise alone with Free Play. The answers are revealed in the game.
The 20 questions
-
Which equation gives the energy needed to change the temperature of a substance?
- E = m L Δθ
- E = c Δθ / m
- E = m / (c Δθ)
- E = m c Δθ
-
What is the SI unit of specific latent heat?
- J m^-3
- J K^-1
- J kg^-1
- J kg^-1 K^-1
-
Specific latent heat of fusion is the energy needed to change the phase of 1 kg of a substance
- without any change in its temperature
- in one second at a fixed power
- while doubling its mass
- while raising its temperature by 1 K
-
A heater of power P runs for time t and heats mass m through a temperature rise Δθ. Which expression gives the specific heat capacity c?
- c = P m / (t Δθ)
- c = P t / (m Δθ)
- c = m Δθ / (P t)
- c = P / (m Δθ t)
-
Why does the temperature of a substance stay constant while it melts?
- Energy goes into breaking bonds, not raising average kinetic energy.
- Energy supplied is converted into sound and light inside the sample, not heat.
- The sample loses as much energy by radiation as it gains from the heater overall.
- The mass of the sample decreases, so its temperature cannot rise further overall.
-
A 2.0 kg block with specific heat capacity 400 J kg^-1 K^-1 is heated from 20 C to 50 C. How much energy is absorbed?
- 24 kJ
- 12 kJ
- 2.4 kJ
- 240 kJ
-
Water has specific heat capacity 4200 J kg^-1 K^-1. How much energy heats 0.50 kg of water by 10 K?
- 2.1 kJ
- 21 kJ
- 42 kJ
- 84 kJ
-
0.25 kg of ice melts completely. The specific latent heat of fusion of ice is 3.34 x 10^5 J kg^-1. How much energy is absorbed?
- 1.34 x 10^6 J
- 8.35 x 10^3 J
- 8.35 x 10^4 J
- 1.34 x 10^5 J
-
A 500 W heater runs for 120 s and heats 2.0 kg of liquid by 15 K with no losses. What is the specific heat capacity of the liquid?
- 1000 J kg^-1 K^-1
- 2000 J kg^-1 K^-1
- 4000 J kg^-1 K^-1
- 8000 J kg^-1 K^-1
-
Energy is supplied to boil 0.10 kg of water already at 100 C, with specific latent heat of vaporisation 2.26 x 10^6 J kg^-1. How much energy is needed?
- 226 kJ
- 22.6 kJ
- 4.52 MJ
- 2.26 MJ
-
A 1.2 kJ energy input heats 0.040 kg of a solid by 25 K. What is its specific heat capacity?
- 4800 J kg^-1 K^-1
- 300 J kg^-1 K^-1
- 120 J kg^-1 K^-1
- 1200 J kg^-1 K^-1
-
A 1.5 kW kettle with no losses heats 1.7 kg of water from 20 C to 100 C, with c = 4200 J kg^-1 K^-1. About how long does it take?
- about 381 s
- about 95 s
- about 857 s
- about 190 s
-
0.30 kg of copper (c = 385 J kg^-1 K^-1) releases 5000 J of energy. By how much does its temperature fall?
- about 130 K
- about 2.3 K
- about 43 K
- about 17 K
-
Energy of 1.0 MJ is supplied to boil water at 100 C, with specific latent heat of vaporisation 2.26 x 10^6 J kg^-1. What mass of water is vaporised?
- about 0.22 kg
- about 2.26 kg
- about 0.44 kg
- about 4.4 kg
-
0.20 kg of ice at 0 C is melted and the water warmed to 30 C. Take L = 3.34 x 10^5 J kg^-1 and c = 4200 J kg^-1 K^-1. What is the total energy needed?
- 67 kJ
- 25 kJ
- 118 kJ
- 92 kJ
-
0.50 kg of water at 80 C is mixed with 0.20 kg of water at 20 C, with no energy lost. What is the final temperature?
- about 63 C
- about 57 C
- about 50 C
- about 70 C
-
Why does an electrical heating experiment often give a calculated specific heat capacity that is too large?
- The thermometer absorbs more energy than the liquid, so the liquid warms more than expected.
- The heater supplies less power than its label, so the energy input is underestimated overall.
- Heat escapes to the surroundings, so the measured temperature rise is smaller than expected.
- Latent heat is added to the liquid, which raises its apparent specific heat capacity too much.
-
Energy to melt 1 kg of ice at 0 C is compared with energy to heat 1 kg of water from 0 C to 10 C (c = 4200 J kg^-1 K^-1, L = 3.34 x 10^5 J kg^-1). What is the ratio of the first to the second?
- about 8.0
- about 0.8
- about 1.3
- about 80
-
A 2.0 kg metal block (c = 500 J kg^-1 K^-1) starts at 15 C and absorbs 60 kJ with no change of phase. What is its final temperature?
- 45 C
- 90 C
- 75 C
- 60 C
-
Which statement about specific heat capacity is correct?
- It is the energy per unit volume needed to melt a substance.
- It is the temperature rise per unit energy supplied to 1 mol of a substance.
- It is the power per unit mass delivered by a heater.
- It is the energy per unit mass needed to raise a substance's temperature by 1 K.
Related quizzes
- Base and derived quantities, SI units and estimation Quiz · 1.1.1 · 20 questions
- Intensity, luminosity and the inverse square law Quiz · 10.1.1 · 20 questions
- Nuclear binding energy and the atomic mass unit Quiz · 11.1.1 · 20 questions
- Gravitational fields and Newton's law of universal gravitation Quiz · 12.1.1 · 20 questions
- Conditions and equations for simple harmonic motion Quiz · 13.1.1 · 20 questions
- Equations for uniformly accelerated motion Quiz · 2.1.1 · 20 questions
- Current, charge and potential difference Quiz · 3.1.1 · 20 questions
- Density and upthrust Quiz · 4.1.1 · 20 questions
- Wave terms and the wave equation Quiz · 5.1.1 · 20 questions
- Impulse and Newton's second law Quiz · 6.1.1 · 20 questions