How do you do the specific heat capacity required practical?

What does the specific heat capacity practical investigate?

It measures the specific heat capacity of a material: the energy needed to raise the temperature of 1 kg of it by 1 °C. You put a measured amount of energy into a block of the material and see how far its temperature rises. The smaller the rise for the energy you put in, the higher the specific heat capacity.

It is a required practical on AQA GCSE Physics and on AQA GCSE Combined Science: Trilogy, so you need it whether you take separate Physics or Combined Science. AQA allows one material or several, so your class might test a single aluminium block or compare it with a copper one.

What equipment do you need?

The core kit is a metal block with two holes drilled in it, one for the heater and one for the thermometer, plus a way to measure the energy you supply.

• A metal block with two holes, often 1 kg of aluminium
• An immersion heater that fits the larger hole, and a low-voltage power supply
• A thermometer or temperature probe for the smaller hole
• A joulemeter, or an ammeter, a voltmeter and a stopwatch
• A top-pan balance to check the mass of the block
• Insulation to wrap round the block, such as cotton wool or bubble wrap
• A pipette and a little water for the thermometer hole
• A heatproof mat

What is the method?

Find the mass of the block, then heat it for about ten minutes while you record the energy supplied and the temperature. The last step is a calculation.

• Measure the mass of the block on the balance and write it down in kilograms.
• Wrap the block in insulation and stand it on the heatproof mat.
• Put the heater in the larger hole. Use the pipette to put a few drops of water in the smaller hole, then put the thermometer in.
• Connect the heater to the power supply through the joulemeter. Without a joulemeter, put the ammeter in series with the heater and the voltmeter across it.
• Record the starting temperature and set the joulemeter to zero.
• Switch on and start the stopwatch. Record the temperature, and the joulemeter reading, every minute for 10 minutes.
• If you are using an ammeter and a voltmeter, write down the current and the potential difference (p.d.). They should stay steady while the heater runs.
• Switch off, but keep watching the thermometer for another minute or so. Record the highest temperature it reaches.

What are the variables?

The energy you put in keeps growing while the heater runs, and the temperature is what you measure as it does. Anything else that could change the temperature rise stays the same. If you compare two metals, the material becomes the independent variable instead, so use blocks of the same mass and the same heater for both.

How do you work out and present the results?

Use ΔE = m c Δθ, rearranged to c = ΔE ÷ (m Δθ). ΔE is the energy transferred in joules, m is the mass in kilograms and Δθ is the temperature rise in °C, so c comes out in J/kg°C. With an ammeter and a voltmeter instead of a joulemeter, find the power first (P = V × I), then the energy (E = P × t, with t in seconds).

Worked example: a heater runs at 12 V and 4.0 A for 10 minutes. Power = 12 × 4.0 = 48 W. Ten minutes is 600 s, so energy = 48 × 600 = 28,800 J. The 1.0 kg aluminium block warms from 20 °C to 50 °C, so Δθ = 30 °C. Then c = 28,800 ÷ (1.0 × 30) = 960 J/kg°C.

The accepted value for aluminium is about 900 J/kg°C, so this result is a little high. That is normal. Some of the energy warms the air and the bench instead of the block, so the temperature rise is smaller than it should be, and a smaller Δθ gives a bigger value of c.

For a graph, plot temperature on the y-axis against energy transferred on the x-axis. The line curves at the start while the heater warms up, then goes straight. The gradient of the straight part is 1 ÷ (m c), so c = 1 ÷ (gradient × m). If you plot temperature against time instead, the gradient is P ÷ (m c), so c = P ÷ (gradient × m).

How do you make it accurate and safe?

Most of the error comes from energy escaping to the surroundings, so insulate the block well, on top as well as round the sides. The drops of water in the thermometer hole matter too. They fill the air gap, so the thermometer reads the temperature of the metal rather than the air in the hole. A longer heating time gives a bigger temperature rise, which makes small errors in reading the thermometer count for less.

Keep reading after you switch off. The heater is still hotter than the block, so energy keeps passing into the metal and the temperature can creep up for a minute or so. Use the highest value you see.

On safety: the heater and the block get hot enough to burn. Don't touch either while the heater is on, and let them cool before you pack away. Keep the water well away from the power supply.

What do exam questions ask about it?

Expect to calculate c from a set of readings, or to rearrange ΔE = m c Δθ to find a mass or a temperature rise. Method questions often ask why the result comes out above the accepted value and how to improve the experiment. Better insulation is the usual answer. You may also be given a graph and asked to use its gradient.

Here is one we wrote in the exam style: "A student's value for the specific heat capacity of copper is higher than the accepted value. Give one reason for this and suggest one improvement to the method. (2 marks)" A good answer names energy lost to the surroundings, then suggests insulating the block.

Common mistakes

• Using the mass in grams. The equation needs kilograms, so a 500 g block is 0.5 kg.
• Using the final temperature instead of the rise. Δθ is the change: final temperature minus starting temperature.
• Timing in minutes. Power × time gives joules only when the time is in seconds.
• Drawing the line of best fit through the curved start of the graph. Use the straight part only.
• Blaming "human error" for a high result. Name the real cause: energy transferred to the surroundings instead of into the block.