AQA GCSE PHYSICS · PAPER 1 · FOUNDATION & HIGHER
Specific heat capacity
Energy · Lesson 3 of 7
Warm-up
Answer each one, then check.
1. What is the unit of temperature change used in this equation?
Degrees Celsius (°C)
2. What is the equation for power?
P = E ÷ t
3. What unit is energy measured in?
Joules
4. Convert 500 g to kilograms.
0.50 kg
5. Which store increases when a substance is heated?
Thermal
Learning Objectives
1. Define specific heat capacity.
2. Apply ΔE = mcΔθ to calculate energy, mass, temperature change or c.
3. Describe the method for Required Practical 1.
4. Explain why measured values are often too high.
5. Interpret energy against temperature graphs.
Specific Heat Capacity
The specific heat capacity of a substance is the energy needed to raise the temperature of 1 kg of the substance by 1 °C.
ΔE = mcΔθ is given on the equation sheet. The unit of c is J/kg °C. For water, c = 4200 J/kg °C.
Required Practical 1: Apparatus
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Insulation reduces energy lost to the surroundings. |
Apparatus to measure specific heat capacity: an insulated metal block with an electric heater, thermometer and power supply.
Method for Required Practical 1
Find the specific heat capacity of a metal block.
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1 Measure the mass Use a balance; record m in kg. |
2 Set up Put the heater and thermometer in the block and wrap it in insulation. |
3 Start Record the starting temperature and switch on the heater. |
4 Record At regular intervals record the temperature and the energy transferred (from the joulemeter, or power times time). |
5 Plot Draw a graph of temperature against energy supplied. |
6 Calculate The gradient is 1/mc, so c = 1/(m × gradient). |
Calculating Energy
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Calculate the energy needed to heat 2.0 kg of water from 20 °C to 35 °C. c = 4200 J/kg °C. |
1. Temperature change
35 − 20 = 15 °C
2. Substitute
ΔE = 2.0 × 4200 × 15
3. Answer
ΔE = 126 000 J
Answer: 126 000 J (126 kJ)
Finding c
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A 0.50 kg metal block needs 9000 J to raise its temperature by 20 °C. Find the specific heat capacity of the metal. |
1. Rearrange
c = ΔE/mΔθ
2. Substitute
c = 9000/(0.50 × 20)
3. Answer
c = 900 J/kg °C
Answer: 900 J/kg °C
Finding the Temperature Change
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3.0 kg of oil (c = 2000 J/kg °C) is given 240 kJ of energy. What is the temperature rise? |
1. Convert energy
240 kJ = 240 000 J
2. Rearrange
Δθ= ΔE/mc
3. Substitute
(240 000)/(3.0 × 2000)
4. Answer
40 °C
Answer: 40 °C
Why the Result Is Too High
A very common exam question.
▸ Energy is lost. Some of the energy from the heater warms the surroundings instead of the block.
▸ So. The energy supplied is more than the energy that actually heats the block.
▸ Effect. c = ΔE/mΔθ uses the energy supplied, so the calculated value is too big.
▸ Fix. Use more insulation, add a lid, or use a lagged block.
Specific Heat Capacity or Latent Heat?
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SPECIFIC HEAT CAPACITY |
SPECIFIC LATENT HEAT |
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▸ Temperature changes. ▸ ΔE = mcΔθ. ▸ Unit: J/kg °C. |
▸ Temperature stays constant (a change of state). ▸ E = mL. ▸ Unit: J/kg. |
Key Terms
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Specific heat capacity Energy needed to raise 1 kg of a substance by 1 °C. |
Thermal energy Energy stored in the moving particles of a substance. |
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Temperature change Final temperature minus starting temperature, Δθ. |
Insulation A material that reduces energy transfer by heating. |
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Joulemeter A meter that shows the energy transferred in joules. |
Gradient How steep a graph line is. |
Your Task: Which Is Hotter?
12 minutes
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1.0 kg of water and 1.0 kg of aluminium (c = 900 J/kg °C) each receive 45 000 J. Calculate the temperature rise of each and say which changes more. 1. Rearrange for temperature change. 2. Compare the two answers. |
A good answer shows: Water: 45 000 ÷ (1.0 × 4200) = 10.7 °C. Aluminium: 45 000 ÷ (1.0 × 900) = 50 °C. Aluminium heats up much more.
Can I...?
☐ Recall the meaning of specific heat capacity.
☐ Use ΔE = mcΔθ.
☐ Rearrange for c.
☐ Convert kJ to J and g to kg.
☐ Describe Required Practical 1.
☐ Explain why the measured c is too high.
☐ Use a graph gradient to find c.
☐ Choose sensible units.
Summary
✓ ΔE = mcΔθ.
✓ The unit of c is J/kg °C.
✓ Measured values of c are too high if energy is lost.
✓ Insulate the block to reduce losses.
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EXAM FOCUS Calculate the energy needed to raise the temperature of 1.5 kg of water by 40 °C. c = 4200 J/kg °C. (2 marks) Substitute all three values then multiply. |