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Specific Heat Capacity Calculations: Q = mcΔT — Answers

Answer sheet with model answers and success criteria for the 10th Grade energy, light, sound & electricity worksheet.

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Success TutoringWorksheetsAnswer sheet10th Grade · Science · Energy, Light, Sound & Electricity

Specific Heat Capacity Calculations: Q = mcΔT — Answers

Part A: Using Q = mcΔT

The specific heat capacity (c) of a substance is the energy needed to raise the temperature of 1 kg of it by 1 °C, in J/kg °C.

Q = mcΔT, where Q is the energy in joules, m is the mass in kg and ΔT is the temperature change in °C.

Specific heat capacities (J/kg °C): water 4200 · aluminum 900 · iron 450 · copper 390

1.How much energy is needed to heat 2 kg of water from 20 °C to 100 °C?

Answer: 672,000 J (672 kJ)

  1. ΔT = 100 − 20 = 80 °C
  2. Q = mcΔT = 2 × 4200 × 80 = 672,000 J
2.A 2000 W kettle supplies this energy. Assuming no energy is wasted, how long does it take to boil the water?

Answer: 336 s (5.6 minutes)

  1. t = E ÷ P
  2. 672,000 ÷ 2000 = 336 s (5.6 min)
3.How much energy is needed to raise the temperature of a 0.5 kg aluminum block by 10 °C?

Answer: 4500 J

  1. Q = 0.5 × 900 × 10 = 4500 J
4.If the same 4500 J were given to a 0.5 kg block of copper, would its temperature rise by more or less than the aluminum's? Calculate it.

Answer: More: about 23.1 °C

  1. ΔT = Q ÷ (mc) = 4500 ÷ (0.5 × 390) = 23.1 °C
  2. Copper has a lower specific heat capacity, so the same energy produces a larger temperature rise
5.Which statement best describes specific heat capacity?
  1. AThe temperature at which a substance boils
  2. BThe energy needed to raise the temperature of 1 kg of a substance by 1 °C
  3. CThe total thermal energy in an object
  4. DHow quickly a substance conducts heat

Answer: The energy needed to raise the temperature of 1 kg of a substance by 1 °C

For water, 4200 J raises 1 kg by 1 °C.

6.A 0.2 kg iron pan cools from 180 °C to 30 °C. How much thermal energy does it give out?

Answer: 13,500 J

  1. ΔT = 180 − 30 = 150 °C
  2. Q = 0.2 × 450 × 150 = 13,500 J

Part B: Rearranging the formula

7.9000 J raises the temperature of a 1 kg metal block by 20 °C. Calculate its specific heat capacity and identify the metal.

Answer: 450 J/kg °C, iron

  1. c = Q ÷ (mΔT) = 9000 ÷ (1 × 20) = 450 J/kg °C
  2. This matches iron
8.21,000 J of energy is given to 0.5 kg of water. By how much does its temperature rise?

Answer: 10 °C

  1. ΔT = Q ÷ (mc) = 21,000 ÷ (0.5 × 4200) = 10 °C
9.84,000 J raises the temperature of some water by 20 °C. What is the mass of the water?

Answer: 1 kg

  1. m = Q ÷ (cΔT) = 84,000 ÷ (4200 × 20) = 1 kg
10.Two objects at the same temperature always contain the same amount of thermal energy.

Answer: False

Thermal energy also depends on the mass and the type of material. A bath of warm water contains far more thermal energy than a cup of water at the same temperature.

Part C: Measuring specific heat capacity

A student placed a 50 W electric heater and a thermometer into holes in a 1.0 kg aluminum block. The heater was switched on for 300 s. The temperature rose from 22 °C to 37 °C.

Energy supplied by a heater: E = P × t (watts × seconds = joules).

11.Calculate the energy supplied by the heater.

Answer: 15,000 J

  1. E = P × t = 50 × 300 = 15,000 J
12.Calculate the specific heat capacity of aluminum from these results.

Answer: 1000 J/kg °C

  1. ΔT = 37 − 22 = 15 °C
  2. c = 15,000 ÷ (1 × 15) = 1000 J/kg °C
13.The accepted value for aluminum is 900 J/kg °C. Explain why the experimental value is too high.

Answer: Some of the heater's energy was lost to the surrounding air and the bench instead of heating the block, so the temperature rise was smaller than it should have been. Dividing by a smaller ΔT gives a larger value of c.

14.Suggest two improvements to make the result more accurate.

Answer: Any two of: wrap the block in insulation; put a lid or insulating base under it; add a drop of oil or water in the thermometer hole for better thermal contact; measure the highest temperature reached after the heater is turned off; repeat and average.

Part D: Apply

15.Water has a very high specific heat capacity. Use this idea to explain two of the following: (a) why water is used in car engine cooling systems and hot-water bottles, (b) why places near the coast have milder temperatures than places far inland, (c) why sand on a beach becomes much hotter than the sea on a sunny day.[4 marks]

Model answer: (a) Water can absorb a lot of energy for only a small rise in temperature, so in an engine cooling system it carries a large amount of heat away from the engine without boiling. In a hot-water bottle, the water stores a lot of energy and releases it slowly as it cools, so it stays warm for a long time. (b) The sea takes a lot of energy to warm up and releases a lot of energy as it cools, so its temperature changes slowly. In summer it keeps nearby land cooler, and in winter it keeps it warmer, giving coastal places milder temperatures. (c) Sand has a much lower specific heat capacity than water, so the same energy from the Sun raises the temperature of the sand much more than that of the sea.

Success criteria:

  • States that water needs a lot of energy to change temperature
  • First chosen situation explained correctly
  • Second chosen situation explained correctly
  • Uses comparison of specific heat capacities or energy storage appropriately
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