Specific Heat Capacity: Why the Time Must Be in Seconds

The question

A student wants to measure the specific heat capacity of aluminium. She places a $50\ \text{W}$ electric heater and a thermometer into holes in an aluminium block of mass $0.75\ \text{kg}$ . The block is wrapped in insulation.

The starting temperature of the block is $19.5^\circ\text{C}$ . She switches on the heater for $6.0$ minutes, and the highest temperature reached by the block is $44.5^\circ\text{C}$ .

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

(b) Assuming all of this energy goes into the block, calculate the specific heat capacity of aluminium. [3 marks]

(c) A data book gives the specific heat capacity of aluminium as $900\ \text{J/(kg}\ {}^\circ\text{C)}$ . Suggest why the student’s value is higher. [1 mark]

Cambridge IGCSE note: the equation for specific heat capacity is Extended content in 0625.

The common wrong answer

Most students start well. They remember $E = Pt$ and put in the numbers from the question:

E = 50 \times 6.0 = 300\ \text{J}

Then they find the temperature change correctly, $44.5 - 19.5 = 25.0^\circ\text{C}$ , and use $c = \dfrac{E}{m\Delta\theta}$ :

c = \frac{300}{0.75 \times 25.0} = 16\ \text{J/(kg}\ {}^\circ\text{C)}

The mistake is in the very first line. A watt is a joule per second, so $E = Pt$ only gives joules when $t$ is in seconds. Using $6.0$ minutes makes the energy, and so the specific heat capacity, exactly 60 times too small. A value of $16\ \text{J/(kg}\ {}^\circ\text{C)}$ would mean aluminium heats up about 260 times more easily than water, which should ring alarm bells.

A second slip I often see is using the final temperature, $44.5^\circ\text{C}$ , instead of the temperature change. That gives about $539\ \text{J/(kg}\ {}^\circ\text{C)}$ , which is also wrong.

The correct working

(a) Convert the time to seconds first:

t = 6.0 \times 60 = 360\ \text{s}

E = Pt = 50 \times 360 = 18\,000\ \text{J}

The energy supplied is $18\,000\ \text{J}$ ( $18\ \text{kJ}$ ).

(b) Temperature change:

\Delta\theta = 44.5 - 19.5 = 25.0^\circ\text{C}

Rearrange $E = mc\Delta\theta$ for $c$ :

c = \frac{E}{m\Delta\theta} = \frac{18\,000}{0.75 \times 25.0} = \frac{18\,000}{18.75} = 960

The specific heat capacity is $960\ \text{J/(kg}\ {}^\circ\text{C)}$ .

(c) Some of the energy from the heater is lost to the surroundings (and some heats the heater and thermometer), so less than $18\,000\ \text{J}$ actually goes into the block. Using the full $18\,000\ \text{J}$ in the calculation makes $c$ come out too high.

Why it works

Power is the rate of transferring energy: $50\ \text{W}$ means $50$ joules every second. To find the total energy you multiply by the number of seconds. In $6.0$ minutes there are $360$ of them, so the heater gives out $18\,000\ \text{J}$ , not $300\ \text{J}$ .

Specific heat capacity is the energy needed to raise the temperature of $1\ \text{kg}$ by $1^\circ\text{C}$ . That’s why the equation uses the change in temperature: the block didn’t warm up from $0^\circ\text{C}$ , so the starting temperature has to come off.

For part (c), it helps to see the numbers. With the data-book value, warming this block by $25.0^\circ\text{C}$ needs $0.75 \times 900 \times 25.0 = 16\,875\ \text{J}$ . The heater supplied $18\,000\ \text{J}$ , so about $1\,125\ \text{J}$ went somewhere other than the block. Insulation reduces that loss but never removes it, which is why this experiment almost always gives a value that is a little too high.

How to spot it next time

Before using $E = Pt$ , check the time unit. If the question says minutes, multiply by 60 first and write the time in seconds on its own line.
Always subtract to get $\Delta\theta$ : final temperature minus starting temperature.
Sense-check your answer. Most metals have specific heat capacities between roughly $100$ and $1\,000\ \text{J/(kg}\ {}^\circ\text{C)}$ , and water is about $4\,200$ . An answer like $16$ means something has gone wrong.