The problems are on the specific heat capacity worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Specific heat capacity is the energy needed to warm 1 kg of a material by 1 °C, and energy transferred equals m × c × Δθ. These worked answers correct the 20 problems of the Physics 9-10 worksheet on specific heat capacity, from meaning of c to a steam burn: each solution shows the formula, the substitution and the final answer with its unit.
Every answer shows the formula, the substitution and the result with its unit. Compare your method as well as your final number, because marks in examinations are awarded for each step. Small rounding differences in the last digit are acceptable.
Level: Physics grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE P3, IGCSE 2.2
Problem 1 Meaning of c
1. It is the energy needed to raise the temperature of 1 kg of a material by 1 °C.2. The unit is J/(kg °C). Δθ is the temperature change, final temperature minus initial temperature.
Problem 2 Warming a pan of water
1. 250 g = 0.25 kg2. ΔE = m c Δθ = 0.25 × 4200 × 12 = 12 600 J = 12.6 kJ
Problem 3 An aluminium pan
ΔE = 2.0 × 900 × 30 = 54 000 J = 5.4 × 10⁴ JProblem 4 Melting ice
E = m L = 0.60 × 334 000 = 200 400 J = 2.0 × 10⁵ J (2 s.f.)Problem 5 Reading a heating curve
1. B: ice melts at 0 °C. D: water boils at 100 °C. The temperature stays constant in both.2. Boiling needs 2.26 × 10⁶ J/kg, about seven times the 3.34 × 10⁵ J/kg for melting, and the energy arrives at the same rate, so it takes about seven times as long.
3. Only section C (0 °C to 100 °C).
Problem 6 A cooling copper block
Δθ = 95 − 25 = 70 °CE = 0.80 × 390 × 70 = 21 840 J = 2.2 × 10⁴ J
Problem 7 Finding the rise
Δθ = E / (m c) = 63 000 / (3.0 × 4200) = 63 000 / 12 600 = 5.0 °CProblem 8 Finding the mass
m = E / (c Δθ) = 8400 / (4200 × 4.0) = 8400 / 16 800 = 0.50 kgProblem 9 Boiling dry
E = 0.25 × 2 260 000 = 565 000 J = 5.65 × 10⁵ JProblem 10 Heater and time
1. ΔE = 0.40 × 4200 × 30 = 50 400 J2. t = E / P = 50 400 / 500 = 100.8 s = 101 s (3 s.f.)
Problem 11 Unknown metal
Δθ = 50 − 18 = 32 °Cc = E / (m Δθ) = 14 400 / (0.50 × 32) = 14 400 / 16 = 900 J/(kg °C), so the metal is aluminium.
Problem 12 Why the sand is hotter
Sea water has a high specific heat capacity, so the same solar energy raises its temperature only a little (and the water also mixes and spreads the energy downward). Sand has a much lower c, so its temperature rises quickly. The hot land warms the air above it, which rises, and cooler air flows in from the sea: a sea breeze.Problem 13 Iron or copper?
1. Iron: 9000 / (1.0 × 450) = 20 °C. Copper: 9000 / (1.0 × 390) = 23.1 °C.2. Copper, because its specific heat capacity is smaller.
Problem 14 Kettle efficiency
1. E = P t = 2400 × 170 = 408 000 J2. ΔE = 1.0 × 4200 × 85 = 357 000 J
3. Efficiency = 357 000 / 408 000 = 0.875 = 87.5 %
Problem 15 Spot the error
The student used the final temperature (40 °C) instead of the temperature change. Δθ = 40 − 15 = 25 °C, so Q = 0.4 × 4200 × 25 = 42 000 J.Problem 16 Ice to steam
Warm ice: 0.20 × 2100 × 10 = 4200 JMelt: 0.20 × 334 000 = 66 800 J
Warm water: 0.20 × 4200 × 100 = 84 000 J
Boil: 0.20 × 2 260 000 = 452 000 J
Total = 4200 + 66 800 + 84 000 + 452 000 = 607 000 J = 6.07 × 10⁵ J
Problem 17 Mixing hot and cool water
Energy lost by hot water = energy gained by cool water.0.30 × 4200 × (80 − T) = 0.50 × 4200 × (T − 20)
0.30 (80 − T) = 0.50 (T − 20)
24 − 0.30 T = 0.50 T − 10
34 = 0.80 T, so T = 42.5 °C
Problem 18 Ice in a drink
Ice melts: 0.030 × 334 000 = 10 020 J. (Cooling the juice to 0 °C would release 25 200 J, so all the ice melts.)Juice loses: 0.25 × 4200 × (24 − T) = 1050 (24 − T)
Melted ice then warms: 0.030 × 4200 × T = 126 T
1050 (24 − T) = 10 020 + 126 T → 25 200 − 1050 T = 10 020 + 126 T
15 180 = 1176 T, so T = 12.9 °C
Problem 19 Wax from a heater trace
1. P lasts 10.2 − 3.0 = 7.2 min = 432 s. E = 250 × 432 = 108 000 J. L = E / m = 108 000 / 0.60 = 1.8 × 10⁵ J/kg.2. Q: rise of 40 °C in 3.0 min = 180 s, so E = 250 × 180 = 45 000 J. c = 45 000 / (0.60 × 40) = 1875 J/(kg °C) (about 1900).
Problem 20 A steam burn
1. Steam: condensing 0.010 × 2 260 000 = 22 600 J, then cooling 0.010 × 4200 × 60 = 2520 J, total 25 120 J (2.5 × 10⁴ J). Water: only 2520 J.2. 25 120 / 2520 ≈ 10 times. Condensation releases a large amount of latent heat on top of the energy released by cooling.
