AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER
Exam Practice: Specific heat capacity
Specific heat capacity · Energy · Lesson 3 of 7 · 18 marks · 19 minutes
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Instructions
• Answer all the questions.
• Write your answers in the spaces provided.
• The marks for each question are shown in brackets - use this as a guide to how much to write.
• The answers are on separate pages at the back. Attempt every question before you look at them.
• Answer all questions. Use the mark allocation as a guide to how much to write.
Question 1 CALCULATE (2 marks)
Calculate the energy needed to raise the temperature of 1.5 kg of water by 40 °C. Specific heat capacity of water = 4200 J/kg °C. Use the equation: change in thermal energy = mass × specific heat capacity × temperature change
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(Total for Question 1 = 2 marks)
Question 2 CALCULATE (4 marks)
A student heats a block of aluminium of mass 0.80 kg using a heater of power 60 W. The graph shows how the temperature of the block changes with time. (a) Calculate the energy transferred by the heater in 300 s. (b) Use the graph and your answer to (a) to calculate the specific heat capacity of aluminium.
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(Total for Question 2 = 4 marks)
Question 3 EXPLAIN (2 marks)
The value of specific heat capacity found in the experiment is higher than the true value. Explain why.
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(Total for Question 3 = 2 marks)
Question 4 DESCRIBE (6 marks)
Describe an investigation to determine the specific heat capacity of an aluminium block. Your answer should include the apparatus, measurements, and how you would use them to obtain the result.
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(Total for Question 4 = 6 marks)
Question 5 EXPLAIN (2 marks)
Water is used in central heating systems. Suggest why.
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(Total for Question 5 = 2 marks)
Question 6 CALCULATE (2 marks)
250 g of water cools from 80 °C to 30 °C. Calculate the energy transferred from the water. Specific heat capacity of water = 4200 J/kg °C.
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(Total for Question 6 = 2 marks)
TOTAL FOR PAPER = 18 MARKS
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Answers and mark scheme
Check your answer only once you have written one.
Question 1 (2 marks)
1.5 × 4200 × 40 = 252 000 J
• Correct substitution 1 mark
• 252 000 J 1 mark
Question 2 (4 marks)
(a) E = Pt = 60 × 300 = 18 000 J. (b) Temperature rise from the graph = 45 − 20 = 25 °C. c = 18 000 ÷ (0.80 × 25) = 900 J/kg °C
• 18 000 J 1 mark
• Temperature rise 25 °C read from graph 1 mark
• c = ΔE ÷ (mΔθ) 1 mark
• 900 J/kg °C 1 mark
Question 3 (2 marks)
Some energy was transferred to the surroundings, so the energy supplied is greater than the energy that heated the block. The equation then gives a larger value of c.
• Some energy is transferred to the surroundings 1 mark
• Energy supplied is more than the energy stored in the block so c appears too high 1 mark
Question 4 (6 marks)
See levels-of-response scheme.
• Level 3 (5 to 6 marks): a clear, ordered method with apparatus (block, heater, thermometer, insulation, power supply or joulemeter), the measurements (mass, temperature, energy or power and time) and the calculation of c with a precaution 5 to 6 marks
• Level 2 (3 to 4 marks): a method with most apparatus and measurements, and some reference to the calculation 3 to 4 marks
• Level 1 (1 to 2 marks): simple statements about heating a block 1 to 2 marks
Question 5 (2 marks)
Water has a high specific heat capacity, so it can store or carry a large amount of energy for each kilogram and each degree of temperature change.
• High specific heat capacity 1 mark
• Can carry or store a lot of energy per kg per degree 1 mark
Question 6 (2 marks)
0.25 × 4200 × 50 = 52 500 J
• Converts to 0.25 kg and uses 50 °C 1 mark
• 52 500 J 1 mark