Specific Heat Capacity Virtual Lab: Aluminium, Copper, Steel and Water | IGCSE Physics

Before you start

You need these ideas and skills for this practical. If any of them are new, read this section first.

Key words

Internal energy: the total energy of all the particles in an object (their kinetic energy and the potential energy of the bonds between them).
Temperature: tells you how fast the particles are moving on average. It is not the same as energy.
Specific heat capacity (c): the energy needed to raise the temperature of 1 kg of a substance by 1 °C.
Immersion heater: an electric heater that fits inside the block (or the water), so that its energy goes straight in.
Insulation: a poor conductor wrapped round the block to reduce the energy lost to the room.
Thermal lag: the heater is hotter than the block, so after it is switched off the block keeps warming up for a short time.

Energy is not temperature

A cup of tea is hotter than a bath of warm water, but the bath stores far more energy because it has far more particles. To warm something up you must give energy to every particle, so:

a bigger mass needs more energy for the same temperature rise;
a bigger temperature rise needs more energy;
a different material needs a different amount of energy: this is its specific heat capacity.

Electrical energy

The heater's power comes from the voltmeter and ammeter readings, and the energy from the time it is switched on:

$$P = VI \qquad \Delta E = Pt = VIt$$

With V in volts, I in amps and t in seconds, ΔE is in joules. A 12 V heater taking 4.17 A has a power of 50 W: it supplies 50 J every second. (Some labs use a joulemeter, which measures ΔE directly.)

Skills you need

Reading a thermometer to the nearest 0.5 °C, with your eye level with the top of the liquid.
Reading digital meters and a stopwatch, and converting minutes to seconds.
Converting grams to kilograms: 1002.9 g = 1.0029 kg.
Plotting a temperature–time graph and finding the gradient of its straight part.

Aim and apparatus

Aim

To determine the specific heat capacity of aluminium by heating a 1 kg block with an electrical heater and measuring the energy supplied and the rise in temperature. Then to compare aluminium with other metals and with water.

Syllabus link: Cambridge IGCSE Physics 0625 / 0972, 2.2.2 Specific heat capacity: know that a rise in temperature increases the internal energy of an object; define specific heat capacity and recall and use c = ΔE/(mΔθ); describe experiments to measure the specific heat capacity of a solid and a liquid.

Apparatus

1 kg block of aluminium with two holes (for the heater and the thermometer)
Insulation (lagging) to wrap round the block, and a heat-proof mat
12 V, 50 W immersion heater and a low-voltage power supply
Ammeter (in series) and voltmeter (across the heater), or a joulemeter
Thermometer, −10 °C to 110 °C, and a drop of oil for the thermometer hole
Stopwatch and a top-pan balance
For water: a beaker with a lid, 250 g of water, and the same heater and thermometer

Theory

Specific heat capacity

$$c = \frac{\Delta E}{m\,\Delta\theta}$$

c = specific heat capacity in J/(kg °C), ΔE = energy supplied in J, m = mass in kg, Δθ = temperature rise in °C. Rearranged: ΔE = mcΔθ.

In this experiment ΔE = VIt and Δθ = θ₂ − θ₁, so:

$$c = \frac{VIt}{m(\theta_2 - \theta_1)}$$

The graph method

While the heater is on, the temperature rises steadily, so a temperature–time graph is a straight line. Its gradient is the rate of temperature rise:

$$\text{gradient} = \frac{\Delta\theta}{\Delta t} = \frac{P}{mc} \quad\Rightarrow\quad c = \frac{P}{m \times \text{gradient}}$$

A steeper line means a smaller specific heat capacity. The graph uses many readings, so it is less affected by one bad reading.

Specific heat capacities

Material	c / J/(kg °C)
Lead	130
Copper	385
Steel	420
Glass	670
Aluminium	910
Air	1000
Ice	2100
Ethanol	2500
Sea water	3900
Water	4200

Values from the Cambridge IGCSE Physics coursebook. Steel's c is about one tenth of water's: the same energy warms 1 kg of steel ten times as much as 1 kg of water.

Why water matters

Water has a very high specific heat capacity. It takes a lot of energy to heat water up, and hot water takes a long time to cool down. That is why:

water is used in central heating and car cooling systems;
the sea warms slowly in summer and cools slowly in winter, so coastal places have milder climates than places far inland;
the land heats up faster than the sea by day, which causes sea breezes.

Planning the investigation

Measurements and variables

Quantity	How
Mass m	Balance, before heating; convert to kg
Voltage V and current I	Voltmeter across the heater, ammeter in series
Heating time t	Stopwatch, started when the heater is switched on
θ₁ and θ₂	Thermometer: θ₁ at the start; θ₂ the highest reading after switching off
Control variables	Same heater and power supply; block insulated; thermometer in oil

How long to heat?

A temperature rise of about 20 °C is a good choice: big enough that the ±0.5 °C reading uncertainty is small compared with Δθ, but not so big that a lot of energy is lost to the room. (The coursebook uses at least 10 °C.)

For the graph method, take a reading every 30 s or every minute.

Method in a real lab

Find the mass m of the block on a balance, in kg. Measure it at the start, before it is heated.
Set up the block on a heat-proof mat. Put the heater in the large hole and the thermometer in the small hole with a drop of oil. Wrap the block in insulation.
Connect the circuit: power supply, ammeter in series with the heater, voltmeter across the heater.
Record θ₁, the starting temperature.
Switch on and start the stopwatch at the same moment. Record V and I. Record the temperature every 30 s.
Switch off after a rise of about 20 °C, and record the time t.
Keep watching the thermometer for a few minutes and record the highest temperature θ₂.
Calculate ΔE = VIt, Δθ = θ₂ − θ₁ and c = ΔE ÷ (mΔθ). Plot temperature against time and use the gradient as a check.
Leave the heater to cool inside the block before you touch it.

How to use this virtual lab

Choose Aluminium, Copper, Steel or Water at the top of the panel.
Step 1, Weigh: press Zero the balance, then put the block on (or pour in the water). The mass is saved.
Step 2, Set up: choose the insulation and oil (both on is best), then press Put in the heater and thermometer.
Step 3, Heat: press Switch on. The meters show V and I, the stopwatch runs, and the temperature is recorded every 30 s. The close-up shows the thermometer scale. Use Speed of time to make the experiment go faster.
Switch off after a rise of about 20 °C. Keep watching: the lab tells you when the temperature has stopped rising and records θ₂.
Step 4, Find c: work out c from the values shown, type your answer and press Check. You will see the book value and why your value is different.
Table & graph: plot temperature against time. Best-fit line finds the gradient of the straight part and the value of c from it; Join points shows the thermal lag after switch-off.
Start a new run to compare: insulation off, no oil, or a different material.

Results

Sample results: aluminium block

m = 1.000 kg, V = 12.0 V, I = 4.17 A, heater switched off at t = 420 s.

t / s	θ / °C	t / s	θ / °C
0	20.0	360	38.0
60	22.5	420 (off)	41.5
120	26.0	480	42.0
180	29.0	540	41.5
240	32.0	600	41.5
300	35.0	660	41.5

The temperature rose slowly at first (the heater was warming up), then steadily. It kept rising after switch-off and reached its highest value, θ₂ = 42.0 °C.

Calculation

$$\Delta E = VIt = 12.0 \times 4.17 \times 420 = 21\,000\ \mathrm{J}$$

$$\Delta\theta = 42.0 - 20.0 = 22.0\ ^\circ\mathrm{C}$$

$$c = \frac{21\,000}{1.000 \times 22.0} = 955\ \mathrm{J/(kg\ ^\circ C)}$$

The book value is 910 J/(kg °C), so this result is about 5% too high.

Analysis and conclusion

The graph

Plot θ (y-axis) against t (x-axis). Draw a best-fit line through the straight part while the heater is on, and find its gradient with a large triangle. From the sample results, (60 s, 22.5 °C) to (420 s, 41.5 °C):

$$\text{gradient} = \frac{41.5 - 22.5}{420 - 60} = 0.0528\ ^\circ\mathrm{C/s}$$

$$c = \frac{P}{m \times \text{gradient}} = \frac{50.0}{1.000 \times 0.0528} = 947\ \mathrm{J/(kg\ ^\circ C)}$$

Why is the value too high?

Almost every class gets a value higher than the book value. The heater supplied ΔE = VIt, but not all of it went into the block:

some was lost to the room (through the insulation and the top of the block);
some heated the heater itself and the thermometer;
for water, some also heated the beaker.

So the energy the block really gained is less than ΔE, and dividing ΔE by mΔθ gives too big a value of c.

Comparing materials

With the same heater for the same time, copper and steel get about twice as hot as aluminium, because their specific heat capacities are about half as big. Water warms up much more slowly than any of the metals. The material with the smallest c has the steepest graph.

Conclusion

“The specific heat capacity of aluminium was found to be 955 J/(kg °C) (947 J/(kg °C) from the graph). This is about 5% higher than the book value of 910 J/(kg °C), because some of the energy supplied by the heater was lost to the surroundings or heated the heater and thermometer, instead of the block.”

Evaluation

Source of error	Effect on c	How to reduce it
Energy lost to the surroundings	Too high: the block gains less energy than VIt	Wrap the block in insulation, including the top; use a smaller temperature rise; start a few degrees below room temperature and finish the same amount above it.
Energy heats the heater and thermometer (and beaker)	Too high	Use a large mass, so this energy is a small fraction of the total.
Poor contact between thermometer and block	Thermometer lags and reads too low, so Δθ is too small and c too high	Put a drop of oil (or water) in the thermometer hole.
Switching off and reading θ₂ straight away	Δθ too small, so c too high	Keep watching after switch-off and record the highest temperature.
Thermometer reading uncertainty ±0.5 °C	A large percentage error if Δθ is small	Use a temperature rise of at least 10–20 °C; use a digital thermometer.
Measuring the mass after heating	The block is hot and may be wet with oil	Measure the mass at the start.
Water not stirred (for water)	Hot water near the heater, cooler water near the thermometer	Stir gently before each reading.

One improvement: use a joulemeter instead of the ammeter, voltmeter and stopwatch, so the energy is measured directly; and add a lid of insulation on top of the block.

Safety

The immersion heater gets extremely hot. Never switch it on unless it is inside the block or the water, and leave it to cool inside the block when you have finished.
Do not touch the hot block: wait for it to cool.
Use a low-voltage supply and keep water away from the power supply and meters.
Handle glass thermometers carefully; report any breakage.
Wash your hands after handling lead or the oil.

Exam practice

Practical-style questions like those in Paper 5 (Practical Test) and Paper 6 (Alternative to Practical), and the questions in the practical book. Try each one before opening the answer.

1. How much energy is needed to raise the temperature of 1.5 kg of water from 20 °C to 100 °C? (c = 4200 J/(kg °C))

ΔE = mcΔθ = 1.5 × 4200 × 80 = 504 000 J (504 kJ).

2. A 12 V heater takes a current of 4.0 A for 5.0 minutes. Calculate the energy it supplies.

t = 5.0 × 60 = 300 s. ΔE = VIt = 12 × 4.0 × 300 = 14 400 J.

3. A 2.0 kg block of metal is heated by 14 400 J and its temperature rises by 18 °C. Calculate its specific heat capacity and suggest what metal it is.

c = ΔE ÷ (mΔθ) = 14 400 ÷ (2.0 × 18) = 400 J/(kg °C). This is between copper (385) and steel (420).

4. Describe how to measure the specific heat capacity of aluminium. Include the measurements, how to make them accurate, and two safety points.

Weigh the block (m). Put the immersion heater and the thermometer (with a drop of oil) in the holes and wrap the block in insulation. Record θ₁. Switch on and start the stopwatch; record V and I. After about a 20 °C rise, switch off and record t. Record the highest temperature θ₂. Calculate c = VIt ÷ (m(θ₂ − θ₁)). For accuracy: insulate the block, use oil for good thermal contact, read the thermometer at eye level, and wait for the highest temperature. Safety: do not touch the hot heater or block; never switch on the heater when it is out of the block; keep water away from electrical equipment.

5. Explain why the block is wrapped in insulation.

To reduce the energy lost to the surroundings. Otherwise some of the energy supplied (VIt) does not go into the block, so the temperature rise is smaller and the calculated value of c is too high.

6. Why does the student keep reading the thermometer after switching the heater off?

The heater is hotter than the block, so energy keeps flowing into the block after switch-off and its temperature keeps rising for a short time. The student must record the highest temperature, or Δθ will be too small and c too large.

7. A student's value for aluminium is 1050 J/(kg °C). The book value is 910 J/(kg °C). Suggest one reason why her value is higher.

Energy was lost to the surroundings (or used to heat the heater and thermometer), so the block gained less energy than VIt. The calculated c is therefore too high.

8. Aluminium (910 J/(kg °C)) or metal X (420 J/(kg °C)) could be used for window frames. Which warms up more on a sunny day? Which would you choose?

Metal X has the smaller specific heat capacity, so the same energy from the Sun raises its temperature about twice as much. Aluminium stays cooler and warms up more slowly, so it is the better choice (and it is also light and does not rust).

9. A 650 W microwave heats a dish of beans from 20 °C to 90 °C. The student calculates c for the beans but notices the dish also got hot. Is her value too high or too low?

Too high. Some of the energy supplied heated the dish, not the beans, so she divided too large an energy by mΔθ.

10. Explain why places near the sea have milder winters than places far inland.

Water has a high specific heat capacity. The sea stores a lot of energy in summer and cools only slowly in winter, so it keeps the nearby land warmer. Land has a low specific heat capacity, so places far inland cool down quickly in winter.