# Cooling corrections

Experiments that involve changing the temperature of a material and measuring that change are necessarily subject to energy transfers between that material or materials and the surrounding environment. These transfers will often not be accounted for and can cause inaccuracies. If the temperatures used are within 10°C or so of the surroundings, the inaccuracy is unlikely to be significant compared to other school laboratory errors. However, if you really want to make the correction, a number of methods can be used, all based upon Newton's law of cooling.

1 In some cases it is possible to cool an object before starting the experiment. You can arrange this so that its temperature difference with the surroundings is equal (but opposite in sign) after heating. It is then reasonable to assume that any thermal transfer away from the object when it is above the temperature of its surroundings is countered by a thermal transfer into the object when its temperature is below. This technique can be employed when mixing liquids, or when measuring the specific thermal capacity of metal blocks.

2 The formal Newton's law method assumes that the rate of loss of heat to the surroundings is proportional to the temperature excess above the surroundings, i.e.

dQ/dt = k(T-Troom)

Where Q is the quantity of energy transferred in a time t

T and Troom are the temperatures of the cooling object and the surroundings respectively,

and k is a constant of proportionality.

Measure the temperature of the object (block, calorimeter, etc.) at the time of start of the heating,t0. Read the temperature at about 30-second intervals until the maximum temperature has been passed and for a significant time after. The longer this time, the more accurate the correction.

Plot the temperature against time on graph paper. On the graph (indicated in diagram below), select times t1 and t3, equal times either side of the maximum temperature at t2. The energy transfer between t2 and t3 is given by integrating the equation above between these values to give:

Q = k ∫(T-Troom)dt

The right-hand side of this equation is proportional to the area under the curve of (T - Troom)versus t, denoted by A2 in the diagram below.

The left-hand side, Q, the heat lost by cooling in the interval ( t3 - t2 ) , is proportional to ΔT3 , the drop in temperature during this time interval.

Remember that Q = mcΔθ, where m is the mass of the cooling body, c is its specific thermal capacity, and Δθ is the drop in temperature.

Therefore ΔT3 = K A2, where K is another constant.

Similarly, the drop in temperature due to cooling in the time interval between t1 and t2, is given by ΔT2 = KA1. (Note that, since the mechanism by which cooling takes place is the same for times between t1 and t2 and between t2 and t3, the constant of proportionality will be the same for both regions.)

So ΔT2 /ΔT3 = A1/A2

If T2 is the temperature observed at time t2, the temperature which the object would have reached had there been no thermal transfer to the surroundings is:

T2 + ΔT2 = T2 + ΔT3 (A1/A2

A1 and A2 can be measured by counting squares on graph paper.

Image courtesy of www.upscale.utoronto.ca/IYearLab/heatcap.pdf

3 If you are using a heater, a simpler method is as follows (courtesy of Frank Grenfell on the CAPT email discussion list).

Observe, (there is no need to record) the temperature as it rises, starting at t0. Turn off the heater and record the time t1. You need this anyway to find the energy input.

Keep the clock running.

Observe the temperature as it continues to rise, and reaches its maximum value (temperatureTmax) at time t2. Keep the clock running.

Record the temperature (T) after another 0.5 t2 (i.e. half as long again as it took to reach the maximum temperature).

The cooling correction to be added is (Tmax - T).

Reasoning. The rate of thermal energy loss while the block is being heated is roughly half what it is at Tmax. So if you observe the temperature drop from Tmax in a time interval equal to half t2, that should be about right.

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