Change of State ยป Practical: Measuring Specific Heat Capacity

What you'll learn this session

Study time: 30 minutes

Edexcel spec: 5.14P

  • How to find the specific heat capacity of water and of a metal block using an electric heater
  • The numbered method, the readings to take and the apparatus to use
  • How to work out the answer from your results
  • Why the answer comes out too high, how to improve it and how to work safely

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The idea: heat something with a known amount of energy

This lesson is examined in Paper 2 only.

In the lesson Specific Heat Capacity you learnt what specific heat capacity means and how to use the equation. Now you measure it. The plan is simple. Give a known mass of material a known amount of energy from an electric heater, then measure how much its temperature rises. You can then work out c.

You need three things: the mass, the energy supplied and the temperature rise.

Key terms:

  • Immersion heater: an electric heater made to be put inside a liquid or a solid block, so that nearly all its heat goes into it.
  • Joulemeter: a meter that measures the energy transferred to a component, in joules, so you do not have to calculate it.
  • Insulation: a material, such as foam or felt, that wraps round the apparatus to slow down the energy escaping.
  • Temperature rise: the final temperature minus the starting temperature.

Finding the energy supplied

There are two ways to find the energy.

🔋 Ammeter, voltmeter and stopwatch

Put an ammeter in series with the heater and a voltmeter across it. Time how long the heater is on. Then use energy = current × voltage × time, E = I × V × t, with the time in seconds.

⚡ Joulemeter

Connect a joulemeter to the heater. It reads the energy in joules directly. This is quicker, but you still need a stopwatch to decide when to stop.

The equation E = I × V × t is on the formulae page of the exam paper, so you do not need to memorise it. The heater is connected to a low voltage supply, usually 12 V.

Method 1: a solid metal block

Apparatus: a metal block of mass 1 kg with two holes in it (for example aluminium), an immersion heater, a thermometer, a low voltage power supply, an ammeter, a voltmeter, a stopwatch, a balance and a layer of insulation.

  1. Measure the mass of the block on the balance and write it down in kg.
  2. Put the heater in one hole and the thermometer in the other. Put a drop of oil in the thermometer hole so that the thermometer is in good contact with the block.
  3. Wrap the block in insulation, such as thick felt or foam.
  4. Record the starting temperature of the block.
  5. Connect the heater to the power supply, with the ammeter in series and the voltmeter across the heater. Switch on and start the stopwatch at the same moment.
  6. Read the current and the voltage and check that they stay the same. Write them down.
  7. After 10 minutes, switch off and stop the stopwatch. Watch the thermometer: it keeps rising for a short while after the heater is off. Record the highest temperature it reaches.
  8. Work out the temperature rise and the energy supplied, then calculate c.

If you use a joulemeter, steps 5 and 6 change: connect it between the supply and the heater and read the energy from the meter at the end.

Method 2: water

Water is a liquid, so the heater goes into a beaker of water instead of a block. The method is similar, with three differences.

  1. Find the mass of the water. Weigh the empty beaker, then weigh it again with the water in it, and subtract.
  2. Use an insulated beaker, or stand the beaker in a larger beaker packed with cotton wool. Put a lid on the top with holes for the heater and the thermometer.
  3. Stir the water gently before each temperature reading. Without stirring, the thermometer only shows the temperature of the water next to it.

The heater must be fully covered by water before it is switched on, and the thermometer bulb must be in the water and not touching the heater or the glass.

Results and a worked calculation

Worked example: aluminium block

QuantityReading
Mass of block1.0 kg
Voltage12 V
Current3.0 A
Time10 minutes = 600 s
Starting temperature18°C
Highest temperature40°C

Step 1: energy supplied = I × V × t = 3.0 × 12 × 600 = 21 600 J

Step 2: temperature rise = 40 − 18 = 22°C

Step 3: c = energy ÷ (mass × temperature rise) = 21 600 ÷ (1.0 × 22) = 982 J/kg°C

To 2 significant figures, c = 980 J/kg°C.

The accepted value for aluminium is 900 J/kg°C, so the answer from the experiment is too high.

Worked example: water

A student heats 0.25 kg of water with a 12 V, 4.0 A heater for 5.0 minutes (300 s). The temperature rises by 11.5°C.

Energy = 4.0 × 12 × 300 = 14 400 J

c = 14 400 ÷ (0.25 × 11.5) = 5009 J/kg°C, so about 5000 J/kg°C.

This is higher than the value for water, 4200 J/kg°C.

Why the answer comes out too high

Some of the energy from the heater is not used to warm the block or the water. It escapes into the air, the table and the insulation. But your calculation counts all the energy supplied. The temperature rise is therefore smaller than it should be.

In c = energy ÷ (mass × temperature rise), a smaller temperature rise on the bottom gives a larger answer. That is why results from this practical are usually higher than the true value. In an exam, say: energy is lost to the surroundings, so the temperature rise is smaller than expected, so c is too high.

Improving the experiment

ImprovementWhy it helps
Insulate the block or the beakerLess energy escapes, so more of it warms the material
Put a lid on the beakerStops energy escaping from the surface of the water, and stops hot water evaporating
Drop of oil in the thermometer holeMakes good contact, so the thermometer reads the block's real temperature
Stir the waterAll the water is at the same temperature, so the reading is correct
Wait for the highest temperatureThe block keeps warming for a moment after the heater is off

Working safely

  • The heater and the block get very hot and stay hot after you switch off. Let them cool before touching them.
  • Never switch on an immersion heater when it is out of the water.
  • Keep the electrical connections and the power supply away from water, and dry your hands.
  • Use a low voltage supply, and switch off before you change any connections.
  • Hot water can scald. Keep the beaker on a stable surface and do not carry it while it is hot.

Common mistakes

Common mistakes

  • Using minutes in E = I × V × t. The time must be in seconds.
  • Using the final temperature instead of the temperature rise.
  • Saying the answer is too high because of "human error". Say that energy is lost to the surroundings.
  • Forgetting to stir the water, so that the thermometer shows the wrong temperature.
  • Reading the thermometer the moment the heater is switched off, when the temperature is still rising.

Exam-style question

A student heats a metal block of mass 1.0 kg with an electric heater connected to a 12 V supply. The current in the heater is 2.5 A and it is switched on for 5.0 minutes. The temperature of the block rises from 21°C to 43°C.

(a) Calculate the energy supplied to the heater. [2 marks]

(b) Calculate the specific heat capacity of the metal. [3 marks]

(c) The student's answer is higher than the true value. Explain why. [2 marks]

(d) Suggest one way to improve the experiment. [1 mark]

Model answer

(a) E = I × V × t = 2.5 × 12 × 300 (1) = 9000 J (1).

(b) Temperature rise = 43 − 21 = 22°C. c = 9000 ÷ (1.0 × 22) (1) = 409 (1) J/kg°C (1). Accept 410.

(c) Some energy is lost to the surroundings (1), so the temperature rise is smaller than it should be, which makes c too high (1).

(d) Wrap the block in insulation (1).

Exam tip

In part (a) the time is given in minutes, so convert it to seconds first. Writing the substitution as well as the answer earns the method mark even if you slip on the arithmetic.

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