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Physics · Energy

Specific heat capacity

Use \(\Delta E = mc\Delta\theta\) to calculate energy changes when a temperature changes, and describe how to determine the specific heat capacity of a material (Required Practical 1).

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Student handouts

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Warm-up

Answer each one, then check.

  1. 1

    What is the unit of temperature change used in this equation?

    Show answerHide answer

    Degrees Celsius (°C)

  2. 2

    What is the equation for power?

    Show answerHide answer

    \(P = E \div t\)

  3. 3

    What unit is energy measured in?

    Show answerHide answer

    Joules

  4. 4

    Convert 500 g to kilograms.

    Show answerHide answer

    0.50 kg

  5. 5

    Which store increases when a substance is heated?

    Show answerHide answer

    Thermal

Learning Objectives

  1. 1Define specific heat capacity.
  2. 2Apply \(\Delta E = mc\Delta\theta\) to calculate energy, mass, temperature change or c.
  3. 3Describe the method for Required Practical 1.
  4. 4Explain why measured values are often too high.
  5. 5Interpret 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.

\(\Delta E = mc\Delta\theta\) is given on the equation sheet. The unit of c is J/kg °C. For water, \(c = 4200\) J/kg °C.

Method for Required Practical 1

Find the specific heat capacity of a metal block.

  1. 1 Measure the mass

    Use a balance; record m in kg.
  2. 2 Set up

    Put the heater and thermometer in the block and wrap it in insulation.
  3. 3 Start

    Record the starting temperature and switch on the heater.
  4. 4 Record

    At regular intervals record the temperature and the energy transferred (from the joulemeter, or power times time).
  5. 5 Plot

    Draw a graph of temperature against energy supplied.
  6. 6 Calculate

    The gradient is \(\dfrac{1}{mc}\), so \(c = \dfrac{1}{m \times \text{gradient}}\).

Calculating Energy

Calculate the energy needed to heat 2.0 kg of water from 20 °C to 35 °C. c = 4200 J/kg °C.

Show the solutionHide the solution
  1. 1 Temperature change \(35 - 20 = 15\) °C
  2. 2 Substitute \(\Delta E = 2.0 \times 4200 \times 15\)
  3. 3 Answer \(\Delta E = 126\,000\) J

Answer\(126\,000\) J (126 kJ)

Finding c

A 0.50 kg metal block needs 9000 J to raise its temperature by 20 °C. Find the specific heat capacity of the metal.

Show the solutionHide the solution
  1. 1 Rearrange \(c = \dfrac{\Delta E}{m\Delta\theta}\)
  2. 2 Substitute \(c = \dfrac{9000}{0.50 \times 20}\)
  3. 3 Answer \(c = 900\) J/kg °C

Answer\(900\) J/kg °C

Finding the Temperature Change

3.0 kg of oil (c = 2000 J/kg °C) is given 240 kJ of energy. What is the temperature rise?

Show the solutionHide the solution
  1. 1 Convert energy \(240\) kJ \(= 240\,000\) J
  2. 2 Rearrange \(\Delta\theta = \dfrac{\Delta E}{mc}\)
  3. 3 Substitute \(\dfrac{240\,000}{3.0 \times 2000}\)
  4. 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

    \(c = \dfrac{\Delta E}{m\Delta\theta}\) 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?

Specific heat capacity

  • Temperature changes.
  • \(\Delta E = mc\Delta\theta\).
  • Unit: J/kg °C.

Specific latent heat

  • Temperature stays constant (a change of state).
  • \(E = mL\).
  • Unit: J/kg.

Which Is Hotter?

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: \(45\,000 \div (1.0 \times 4200) = 10.7\) °C. Aluminium: \(45\,000 \div (1.0 \times 900) = 50\) °C. Aluminium heats up much more.

Can I...?

  1. 1Recall the meaning of specific heat capacity.
  2. 2Use \(\Delta E = mc\Delta\theta\).
  3. 3Rearrange for c.
  4. 4Convert kJ to J and g to kg.
  5. 5Describe Required Practical 1.
  6. 6Explain why the measured c is too high.
  7. 7Use a graph gradient to find c.
  8. 8Choose sensible units.

Summary & Exam Focus

  • \(\Delta E = mc\Delta\theta\).
  • 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.

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) (2 marks)

Substitute all three values then multiply.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

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.
Temperature change
Final temperature minus starting temperature, \(\Delta\theta\).
Insulation
A material that reduces energy transfer by heating.
Joulemeter
A meter that shows the energy transferred in joules.
Gradient
How steep a graph line is.

Questions and answers

11 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Calculate 2 marks Easier

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

Mark scheme — 2 marks available

  • Correct substitution — 1 mark
  • 252 000 J — 1 mark

Model answer

\(1.5 \times 4200 \times 40 = 252\,000\) J

2. Exam question Calculate 4 marks Easier

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.

A straight line graph of temperature against time rising from 20 degrees Celsius to 45 degrees Celsius over 300 seconds.

Mark scheme — 4 marks available

  • 18 000 J — 1 mark
  • Temperature rise 25 °C read from graph — 1 mark
  • \(c = \Delta E \div (m\Delta\theta)\) — 1 mark
  • 900 J/kg °C — 1 mark

Model answer

(a) \(E = Pt = 60 \times 300 = 18\,000\) J. (b) Temperature rise from the graph = 45 − 20 = 25 °C. \(c = 18\,000 \div (0.80 \times 25) = 900\) J/kg °C

3. Exam question Explain 2 marks Easier

The value of specific heat capacity found in the experiment is higher than the true value. Explain why.

Mark scheme — 2 marks available

  • 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

Model answer

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.

4. Exam question Describe 6 marks Core

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.

Mark scheme — 6 marks available

  • 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

Model answer

See levels-of-response scheme.

5. Exam question Explain 2 marks Easier

Water is used in central heating systems. Suggest why.

Mark scheme — 2 marks available

  • High specific heat capacity — 1 mark
  • Can carry or store a lot of energy per kg per degree — 1 mark

Model answer

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.

6. Exam question Calculate 2 marks Easier

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.

Mark scheme — 2 marks available

  • Converts to 0.25 kg and uses 50 °C — 1 mark
  • 52 500 J — 1 mark

Model answer

\(0.25 \times 4200 \times 50 = 52\,500\) J

7. Multiple choice 1 mark Easier

The unit of specific heat capacity is...

  1. A J/kg
  2. B J/°C
  3. C kg °C/J
  4. D J/kg °C Correct

Why: Joules per kilogram per degree Celsius.

8. Multiple choice 1 mark Core

A larger specific heat capacity means the material...

  1. A needs more energy to warm up Correct
  2. B warms up quickly
  3. C cannot be heated
  4. D has a lower mass

Why: It needs more energy per kilogram for each degree.

9. Multiple choice 1 mark Core

To raise 2 kg by 10 °C when c = 500 J/kg °C needs...

  1. A 1000 J
  2. B 10 000 J Correct
  3. C 5000 J
  4. D 500 J

Why: \(2 \times 500 \times 10 = 10\,000\) J.

10. Multiple choice 1 mark Core

Insulating the block in the experiment...

  1. A increases the mass
  2. B increases c
  3. C reduces energy lost to the surroundings Correct
  4. D removes the thermometer error

Why: Less energy is lost, so results are more accurate.

11. Multiple choice 1 mark Stretch

Water has c = 4200 J/kg °C. Compared with a metal of c = 400 J/kg °C, it...

  1. A heats up faster
  2. B has less mass
  3. C needs less energy
  4. D needs more energy for the same rise Correct

Why: Higher c means more energy per kg per °C.