Specific Heat Capacity and Energy Stores

Two ways are suggested in GCSE Physics for measuring the specific heat capacity of a material. I have already explained the electrical method (see here) but it looks likely, from the advanced information supplied by AQA, that the other method will feature in this summer’s Trilogy physics examination. So here’s what you need to know about the alternative approach.

Whereas the electrical method uses energy transfers, the alternative approach is based on energy stores. The basic idea is as follows;

• heat a metal block of known mass to a known temperature (this can be done by placing the block into hot water)
• move the hot metal block to a glass beaker containing a known volume of cold water (at a known temperature)
• measure the maximum temperature achieved by the water

So what is going on here?

The hot metal block has a store of thermal energy and when it is placed in cold water the energy starts to transfer away from the block, increasing the temperature of the water while at the same time reducing the temperature of the metal block.

This process will continue until thermal equilibrium is achieved and the two temperatures are the same. Obviously, I hope, that will coincide with the water reaching its maximum temperature – before everything eventually cools down to the temperature of the surroundings.

As well as being asked to describe this experiment, you could be asked to calculate various steps in this process.

You might be told that the metal block has a mass of 500 g and that it is heated to 84 °C. You would also be given the specific heat capacity of the metal – let’s say it’s iron, so the specific heat capacity is 440 J / kg °C. (You should recall this means that 440 J is required to raise the temperature of 1 kg of iron by 1 °C.) You are then asked to calculate how much energy is transferred (away from the iron block) when it cools to 26 °C, which was the maximum temperature reached by the cold water.

Either alongside the question or on the Equation Sheet you will find the correct relationship that is needed to solve this problem:

ΔE = m c Δθ
change in energy = mass x specific heat capacity x change in temperature
(J) (kg) (J / kg °C) (°C)

Here’s how to use the equation and calculate the required answer…

• First, convert 500 g to 0.500 kg so the units for mass match the units for specific heat capacity.
• Next, calculate the temperature change for the iron block, which is 58 °C (84 – 26).
• Finally, substitute these numbers, together with the value you were given for the specific heat capacity of iron (440 J / kg °C) into the equation and calculate the result.

You should get a final answer of 12 760 J.

For the second part of the question (probably only found in a Higher Tier paper) you could be asked to calculate the specific heat capacity of water assuming that all the thermal energy stored in the iron black is transferred to the water. You would need to know the mass of water used (let’s say it was 0.400 kg) and that it started at room temperature (20 °C).

First you need to rearrange the equation so that specific heat capacity is the subject (I’ll leave you to do that) then substitute the values to calculate the answer required. But wait! What value do you use for the temperature change? Not 58 °C, as in the first part of the question: the temperature change for the water was much less; only 6 °C (26 – 20). And to add an extra mark, you must give your answer to three significant figures.

You should get a final answer of 5320 J / kg °C (not 5317, because that would be four significant figures).

To round-off the question, you might be asked to explain why the calculated value is higher than the textbook value of 4200 J / kg °C. Think about this…

If you do the calculation using the textbook specific heat value and assume that all of the iron block’s energy is transferred to the water then the temperature rise for the water would have been greater (see footnote). That might suggest some heat energy was lost to the surroundings, which is probably an acceptable answer but I wouldn’t allow it because, in this case, “the surroundings” is too just vague.

I would prefer you to state one of the two following reasons;

• the iron block cooled slightly by transferring energy to the air when it was being moved to the beaker of cold water, so the temperature of the iron block when it entered the cold water was lower than expected.
• the iron block is in contact with the glass beaker when submerged in the cold water, so the beaker would have taken some of the heat from the iron block and not all of the energy would have transferred to the water.

Note that calculating a value of specific heat capacity that is too high means some energy must have been lost elsewhere in the experiment.

Footnote: For those of you who have got this far, here are two more things to think about;

Firstly (Foundation Tier), why is it important that the iron block is heated to a temperature less than 100 °C? (HINT: What happens to water at this temperature?)

Secondly (Higher Tier), if you calculated the temperature rise for the water by assuming that all the energy stored in the hot iron block is transferred to the water, using the textbook value for the specific heat capacity of water (4200 J / kg °C) you would have found the maximum water temperature would have been 27.6 °C – but this cannot be correct. Why? (HINT: Think about the tempearture difference used to calculate the energy stored in the iron.)

If you want to email me your answers for feedback then please do so using mrtarrant (usual symbol) cantab.net