🔋 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.
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Unlock This CourseThis 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:
There are two ways to find the energy.
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.
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.
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.
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.
Water is a liquid, so the heater goes into a beaker of water instead of a block. The method is similar, with three differences.
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.
| Quantity | Reading |
|---|---|
| Mass of block | 1.0 kg |
| Voltage | 12 V |
| Current | 3.0 A |
| Time | 10 minutes = 600 s |
| Starting temperature | 18°C |
| Highest temperature | 40°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.
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.
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.
| Improvement | Why it helps |
|---|---|
| Insulate the block or the beaker | Less energy escapes, so more of it warms the material |
| Put a lid on the beaker | Stops energy escaping from the surface of the water, and stops hot water evaporating |
| Drop of oil in the thermometer hole | Makes good contact, so the thermometer reads the block's real temperature |
| Stir the water | All the water is at the same temperature, so the reading is correct |
| Wait for the highest temperature | The block keeps warming for a moment after the heater is off |
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]
(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).
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.