OpenRevise GCSE

Viewing as

Student Teacher

Teaching this? The teacher view adds 2 files, the mark schemes and the model answers.

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).

  • 6 key terms
  • All boards
Download the full pack · 3 files

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.

Practice questions

Have a go at each one before you open its answer.

  1. 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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • Correct substitution — 1 mark
    • 252 000 J — 1 mark
  2. 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.

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

    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

    Mark scheme

    • 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
  3. Question 3 Explain 2 marks

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

    Show answerHide answer

    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.

    Mark scheme

    • 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
  4. 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.

    Show answerHide answer

    Model answer

    See levels-of-response scheme.

    Mark 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
  5. Question 5 Explain 2 marks

    Water is used in central heating systems. Suggest why.

    Show answerHide answer

    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.

    Mark scheme

    • High specific heat capacity — 1 mark
    • Can carry or store a lot of energy per kg per degree — 1 mark
  6. 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.

    Show answerHide answer

    Model answer

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

    Mark scheme

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

Quick check

  1. The unit of specific heat capacity is...

    1. AJ/kg
    2. BJ/°C
    3. Ckg °C/J
    4. DJ/kg °C
    Show answerHide answer

    D: J/kg °C

    Joules per kilogram per degree Celsius.

  2. A larger specific heat capacity means the material...

    1. Aneeds more energy to warm up
    2. Bwarms up quickly
    3. Ccannot be heated
    4. Dhas a lower mass
    Show answerHide answer

    A: needs more energy to warm up

    It needs more energy per kilogram for each degree.

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

    1. A1000 J
    2. B10 000 J
    3. C5000 J
    4. D500 J
    Show answerHide answer

    B: 10 000 J

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

  4. Insulating the block in the experiment...

    1. Aincreases the mass
    2. Bincreases c
    3. Creduces energy lost to the surroundings
    4. Dremoves the thermometer error
    Show answerHide answer

    C: reduces energy lost to the surroundings

    Less energy is lost, so results are more accurate.

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

    1. Aheats up faster
    2. Bhas less mass
    3. Cneeds less energy
    4. Dneeds more energy for the same rise
    Show answerHide answer

    D: needs more energy for the same rise

    Higher c means more energy per kg per °C.

Downloads

Free to keep, print and annotate.

Download all 3 as a zip
Something here looks wrong?

Tell us what and we will go and look. It goes to whoever writes these pages, nobody else, and we do not ask who you are — so there is nothing to sign and nothing comes back to you.