3.5 Required practical: specific heat capacity
9 exam questions worth 22 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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The block is wrapped in insulation. Explain why.
Mark scheme
reduces the energy transferred to the surroundings / less thermal energy lost (1); so more of the energy supplied heats the block / the result is more accurate (1)
Model answer
Insulation reduces the energy transferred to the surroundings, so more of the energy supplied goes into heating the block and the result is more accurate.
A little oil is put in the hole for the thermometer. Explain why.
Mark scheme
to improve thermal contact between the thermometer and the block (1); so the thermometer measures the temperature of the block accurately (1)
Model answer
The oil improves the thermal contact between the block and the thermometer, so it measures the temperature of the block accurately.
A heater with a potential difference of 12 V and a current of 4.0 A is on for 5.0 minutes. Calculate the energy it transfers.
Mark scheme
t = 5.0 × 60 = 300 s (1); E = V I t = 12 × 4.0 × 300 (1); 14 400 J (1)
Model answer
t = 300 s. E = V I t = 12 × 4.0 × 300 = 14 400 J.
The heater transfers 14 400 J to a 1.0 kg metal block and its temperature rises by 16 °C. Calculate the specific heat capacity of the metal.
Mark scheme
c = ΔE / (m Δθ) (1); 14 400 / (1.0 × 16) (1); 900 J/kg °C (1)
Model answer
c = ΔE / (m Δθ) = 14 400 / (1.0 × 16) = 900 J/kg °C.
A student's result for the specific heat capacity is higher than the accepted value. Suggest a reason and explain how it causes the result to be too high.
Mark scheme
some energy was transferred to the surroundings / lost by heating (1); so the temperature rise of the block was smaller than expected, making c = ΔE / (m Δθ) too large (1)
Model answer
Some of the energy supplied was transferred to the surroundings. The temperature rise of the block was smaller than expected, so c = ΔE / (m Δθ) is too high.
Which equation gives the electrical energy transferred by the heater?
Choose one answer.
Correct Not quite: the answer is E = V I t
Mark scheme
A: E = V I t
Model answer
E = V I t (A): the energy is the potential difference multiplied by the current and the time.
In an experiment, a 1.0 kg block of metal is heated and its temperature rises from 20.0 °C to 60.0 °C when 36 000 J is supplied. Calculate the specific heat capacity of the metal.
Mark scheme
Δθ = 40.0 °C (1); c = ΔE ÷ (m Δθ) = 36 000 ÷ (1.0 × 40.0) (1); 900 J/kg °C (1)
Model answer
Δθ = 40.0 °C. c = 36 000 / (1.0 × 40.0) = 900 J/kg °C.
A student finds that the specific heat capacity of aluminium is 1020 J/kg °C. The accepted value is 900 J/kg °C. Calculate the percentage difference.
Mark scheme
(1020 − 900) ÷ 900 × 100 (1); 13% (1)
Model answer
Percentage difference = (1020 − 900) / 900 × 100 = 13%.
Explain two ways in which the student could reduce the energy that is wasted in this experiment.
Mark scheme
wrap the block in insulation (1); to reduce the energy transferred to the surroundings (1); put oil in the thermometer hole / insert the heater fully (1); so there is good thermal contact and the temperature measured is more accurate (1)
Model answer
Wrap insulation around the block, so less energy is transferred to the surroundings. Put a little oil in the thermometer hole, so there is good thermal contact and the thermometer reads the block's temperature accurately.
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