P2 RP Required Practical: Specific Heat Capacity

A free GCSE Combined Science: Trilogy lesson for AQA (8464) from OBA School: measuring the specific heat capacity of a metal block with an electric heater, a joulemeter (or ammeter, voltmeter and stopclock) and a thermometer. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

• Describe how to measure the specific heat capacity of a metal block, including the apparatus and circuit
• Calculate the energy supplied using a joulemeter reading or E = V I t
• Calculate specific heat capacity from results, including from the gradient of a temperature–time graph
• Identify the variables and hazards, and explain the main sources of error, how to reduce them and how resolution affects precision

Key words

• Joulemeter: A meter that measures the energy transferred by the electricity supply to the heater, in joules.
• Immersion heater: A small electric heater that fits tightly in a hole in the block.
• Thermal contact: How well two objects touch so energy can pass between them. A drop of oil or water in the thermometer hole improves it.
• Gradient: The slope of a straight-line graph: change in y ÷ change in x. Here, temperature rise per second.
• Systematic error: An error that shifts every result the same way, e.g. energy lost to the surroundings makes c too big every time.
• Random error: Unpredictable scatter in readings, reduced by repeating and taking a mean.

Aim, apparatus and circuit

Aim: to determine the specific heat capacity of a metal block by measuring the energy transferred to it and its temperature rise.

Without a joulemeter, use an ammeter in series with the heater, a voltmeter in parallel across it and a stopclock. Then the power is P = V I and the energy supplied is E = P t = V I t.

Method

1. Measure the mass of the block with a top-pan balance (e.g. a 1 kg aluminium block).

2. Put the heater in the large hole and the thermometer in the small hole. Add a drop of oil to the thermometer hole.

3. Wrap the block in insulation and stand it on a heatproof mat.

4. Record the starting temperature and reset the joulemeter (or note the voltmeter and ammeter readings).

5. Switch on and start the stopclock. Record the temperature (and joulemeter reading) every minute for 10 minutes.

6. Switch off. Keep watching the thermometer and record the highest temperature it reaches.

7. Calculate the energy supplied and the temperature rise, then c = ΔE ÷ (m Δθ).

Independent variable: time of heating (energy supplied)

Dependent variable: temperature of the block

Control variables: mass of the block; same heater and power supply setting (same p.d.); same insulation; same starting temperature

Safety: Hazards: the heater and block get hot enough to burn – do not touch them while on and let them cool before packing away. Use a low-voltage supply and keep water away from the electrical connections.

Processing the results

Worked example: using totals

A 1.00 kg aluminium block. Heater: 12.0 V, 3.5 A for 10 minutes. Temperature rose from 18.0 °C to 41.9 °C (graph below).
P = V I = 12.0 × 3.5 = 42 W; t = 600 s; E = P t = 42 × 600 = 25 200 J
Δθ = 41.9 − 18.0 = 23.9 °C
c = ΔE ÷ (m Δθ) = 25 200 ÷ (1.00 × 23.9) = 1054 J/kg °C

Answer: about 1050 J/kg °C (accepted value about 900 J/kg °C)

Using a graph. Plot temperature (y) against time (x). After a short lag, the graph is a straight line. Its gradient is the rate of temperature rise, Δθ/Δt. Since P t = m c Δθ:

c = P ÷ (m × gradient) (gradient in °C per second, P in watts, m in kg)

Worked example: from the gradient

gradient = 16.2 ÷ 360 = 0.045 °C per second
c = 42 ÷ (1.00 × 0.045) = 933 J/kg °C

Answer: about 930 J/kg °C – closer to the accepted value, but still too high because energy is lost to the surroundings

Using the gradient ignores the slow start (the lag while the heater and the metal near it warm up), so it is usually better than using just the first and last readings.

Errors and improvements

• Energy is lost to the surroundings; Effect on the value of c Δθ is too small, so c is too large (systematic error); Improvement insulate the block well; lag the top too
• Some energy heats the heater and thermometer, not the block; Effect on the value of c c too large; Improvement use a heater that fits tightly; this effect is small
• Thermometer lags behind the block; Effect on the value of c final temperature too low if read at switch-off; Improvement record the highest temperature after switching off
• Poor thermal contact in the thermometer hole; Effect on the value of c readings too low; Improvement add a drop of oil or water
• Thermometer resolution 1 °C; Effect on the value of c large uncertainty in small rises; Improvement use a digital thermometer (0.1 °C) and heat for longer

Exam tip: If you are asked why your value is higher than the accepted value, the answer is almost always: “energy was transferred to the surroundings, so the temperature rise was smaller than it should have been.”

Common mistakes

• Saying energy losses make c smaller. Losses make the temperature rise smaller, so the calculated c is larger.
• Using time in minutes in E = V I t. Time must be in seconds.
• Putting the voltmeter in series or the ammeter in parallel.
• Using the whole curve, including the lag, to work out the gradient.

Revision questions

Try these, then check your answers below each one.

Question 1. The diagram shows the apparatus for measuring the specific heat capacity of a metal block. Which label shows the joulemeter? (Choose from: A, B, C, D.)

Answer: C

Working: C is the meter wired between the supply and the heater: the joulemeter. A is the metal block, B the immersion heater and D the thermometer.

Question 2. A heater runs at 11.5 V and 4.0 A for 8 minutes. Calculate the energy it supplies.

Answer: 22080 J

Working: t = 8 × 60 = 480 s. E = V I t = 11.5 × 4.0 × 480 = 22 080 J.

Question 3. True or false? Energy lost to the surroundings makes the calculated specific heat capacity bigger than the true value.

Answer: True

Working: True. The temperature rise is smaller than it should be, so c = ΔE ÷ (m Δθ) comes out too big.

Related lessons

• Previous lesson: P2.3 Heating and Insulating Buildings
• Next lesson: P3.1 Energy Demands
• In this chapter: P2.1 Energy Transfer by Conduction
• In this chapter: P2.2 Specific Heat Capacity

Continue in the full course

This lesson is section P2 RP (AGCSC3104) of our free GCSE Combined Science course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Combined Science course.