DETERMINATION OF THE SPECIFIC HEAT OF WATER. 445 
The rate of cooling can only be affected by evaporation in so far as it increases with 
the temperature, and since for small changes it would vary as a linear function of the 
temperature, any error due to evaporisation is eliminated by the cooling correction. 
Some Corrections to the Water-equivalents. 
1. Correction for Surrounding Air.—The usual method of finding the value of 
the correction depends on the assumption that the loss of heat is proportional to the 
excess of temperature, and as regards radiation this will hold whenever the variation 
is small. When however an appreciable quantity of heat is carried away by 
conduction, an error may be introduced by taking as the excess of temperature the 
difference observed between that inside the calorimeter and that of its jacket, for 
the temperature gradient in that case will not be uniform during the whole course of 
the experiment. Consider, for instance, a calorimeter placed on a more or less 
conducting material like wood or cork. If the calorimeter is suddenly heated, 
the temperature gradient at the surface will be large and will then slowly diminish. 
The cooling observed, as usual, after the lapse of a few minutes, will underrate the 
heat lost during the first few minutes. It is easy to see that when conduction 
comes into play, the amount of heat lost or gained will depend on the distribution of 
heat in the conducting body, and that will depend on the previous history of the 
calorimeter. Thus, when a calorimeter is kept heated until all the surrounding 
parts have reached a steady temperature, and observations are then made on the 
rate of cooling, we are not justified in applying the rate thus formed to the period in 
which heat is communicated to the calorimeter. An estimate may be made of the 
error introduced if the correction is applied in the usual way. We take the case of 
a calorimeter placed on a non-conducting slab of thickness c. At the beginning of 
the experiment everything is to be at temperature zero. At the time ¢ = 0, the 
temperature (%) of the calorimeter is to be gradually and uniformly raised so 
that we may take wu = pt. Finally, at a time T, the temperature is kept constant 
and equal to pT. It is required to find the rate at which heat is given up by the 
calorimeter at any moment, and the total amount of heat which passed through its 
surface. 
Taking the flow to be linear, the differential equation to be satisfied is 
du/ot = a? 0*u/dx?. 
Here a = x/s, where « represents the conductivity and s the heat capacity per unit 
volume. We introduce the conditions 
yu = 0 if t = 0 for all values of a. 
— ¢ (t) for 7 = 0. 
u = 0 for x = ¢ for all values of t. 
