Specific heat capacity

The specific heat capacity of a material is the energy needed to warm 1 kg of it by 1°C. Water needs a lot, 4,200 J for each kilogram and degree, while metals need much less. The energy to change the temperature of anything is E = mcΔT, and you can measure c by heating a block with an electric heater.

Key facts

  • Energy to change temperature: E = mcΔT (c = specific heat capacity, J/kg °C).
  • Water: c = 4,200 J/kg °C, much higher than most substances.
  • To measure c, heat a block with an electric heater: c = E ÷ (mΔT). Energy lost to the surroundings makes the result too high.
  • When hot and cold mix: energy lost by the hot = energy gained by the cold.

Worked examples

Example 1

A cup of milk is heated from 17°C to 60°C. What is the temperature change?

  1. It starts at 17°C and ends at 60°C.
  2. Temperature change = final − starting temperature.
  3. 60 − 17 = 43°C

Answer: 43 °C

Example 2

How much energy heats 2.0 kg of water from 20°C to 70°C?

  1. Temperature change: ΔT = 70 − 20 = 50°C
  2. Energy = mass × specific heat capacity × temperature change. E = 2.0 × 4,200 × 50 = 420,000 J

Answer: 420,000 J

Example 3

What temperature rise does 21,000 J of energy give 0.50 kg of water? (c = 4,200 J/kg°C)

  1. ΔT = E ÷ (m × c)
  2. ΔT = 21,000 ÷ (0.50 × 4,200) = 21,000 ÷ 2,100 = 10°C

Answer: 10°C

Example 4

1.0 kg of water and 1.0 kg of copper are each given 4,000 J. Which warms up more, and by how much? (water: 4,200 J/kg°C; copper: 385 J/kg°C)

  1. Water: ΔT = 4,000 ÷ (1.0 × 4,200) ≈ 0.95°C
  2. Copper: ΔT = 4,000 ÷ (1.0 × 385) ≈ 10.4°C
  3. Copper has a much lower specific heat capacity, so the same energy warms it far more.

Answer: the copper: about 10.4°C, against about 0.95°C for the water

Example 5

A 1.5 kW heater warms a 2.0 kg aluminium block for 2 minutes. The temperature rises by 90°C. Find c for aluminium (assume no losses).

  1. Energy from the heater: E = Pt = 1,500 × 120 = 180,000 J
  2. Rearrange E = mcΔT. c = 180,000 ÷ (2.0 × 90) = 1,000 J/(kg °C)
  3. The accepted value is about 900 J/(kg °C). Heat lost to the room makes our value too high.

Answer: 1,000 J/(kg °C)

Example 6

A 50 W heater warms a 1.0 kg aluminium block for 300 s, and its temperature rises by 16°C. Find the specific heat capacity it measures. Why is it more than the true value, 900 J/kg°C?

  1. Energy supplied: E = P × t = 50 × 300 = 15,000 J
  2. c = 15,000 ÷ (1.0 × 16) ≈ 940 J/kg°C
  3. Some energy escapes to the air, so the block warms less than it should and c comes out too big. Insulating the block helps.

Answer: about 940 J/kg°C; some energy heats the surroundings instead of the block

Example 7

A student measures the specific heat capacity of aluminium as 1,050 J/kg °C, higher than the true value of 900. What is the most likely reason?

  1. c = E ÷ (mΔT), using the energy supplied by the heater.
  2. Some of that energy heats the surroundings, not the block.
  3. So the measured energy is too big, giving a value of c that is too high.

Answer: energy was lost to the surroundings, so more energy was measured than actually heated the block

Example 8

A 3,000 W kettle is 93% efficient. How long does it take to heat 1.1 kg of water from 16°C to 100°C? Give your answer to the nearest second. (c of water = 4,200 J/kg °C.)

  1. E = 1.1 × 4,200 × 84 = 388,080 J
  2. Useful power: 0.93 × 3,000 = 2,790 W
  3. t = 139.1 s

Answer: 139 s

Example 9

0.5 kg of water at 83°C is mixed with 0.7 kg of water at 16°C in an insulated jug. What is the final temperature, to 1 decimal place?

  1. Energy lost by the hot water = energy gained by the cold water; c cancels.
  2. 0.5(83 − T) = 0.7(T − 16)
  3. T = (0.5 × 83 + 0.7 × 16) ÷ 1.2 = 43.92°C

Answer: 43.9°C