There are several simple methods for measuring the specific heat capacities of both solids and liquids, such as the method of mixtures, but we will consider here only electrical methods. (You will find a consideration of the method of mixtures in the example in Heat energy 1). Since the specific heat capacity varies with temperature, we have seen it is important to record the mean temperature at which the measurement is made.
qo and q1 are the initial and final temperatures of the specimen and q is the heat loss. Using the
cooling correction, the value of q may be found. This simple method can be used for liquids or
solids, although in the case of a liquid allowance has to be made for the thermal capacity of the
container, and the liquid should also be stirred to allow an even distribution of the heat energy
throughout its volume. This is necessary since liquids are such poor conductors
This was first developed by Callender
and Barnes in 1902 for the measurement of the specific heat capacity of a Liquid, and is shown
in Figure 2. Its main advantage is that the thermal capacity of the apparatus itself need not be
Liquid flows in
from a constant-head apparatus at a constant rate past a thermometer (θ>o). It then flows around the heater coil and out past a second
thermometer where the outlet temperature (θ1) may be
measured. When steady-state conditions have been reached (a temperature difference between
inlet and outlet points of 5 oC is reasonable) the temperatures and the flow rate of the
liquid are measured. A vacuum jacket round the heater coil reduces heat losses. The
electrical energy supplied to the heater coil (E = VIt) may be found readily with a joulemeter or
with an ammeter and voltmeter.
Two sets of measurements are carried out.
For a first experiment we have:
Electrical energy supplied (E1) = V1I1t1 = m1c(θ1 - qo ) + H
where c is the specific heat capacity of the liquid and the heat loss to the surroundings and to the apparatus.
The flow rate and rate of energy input are now altered to give a second set of results. However, if the inlet and outlet temperatures are the same as in the first experiment the heat loss will also be the same. Therefore:
Electrical energy supplied (E2) = V2I2t2 = m2c(θ1 - θo ) + H
Eliminating the heat loss (H) gives: