164 Domestic Science 



106. In actual practice it is, of course, almost 

 impossible to prevent the loss of some of the heat 

 supplied by the hot body, and, in any case, when a 

 liquid is heated by it, a portion of the heat is used 

 in raising the temperature of the containing vessel. 

 Allowance must, therefore, be made for this in calcula- 

 tion. This is usually done by finding the thermal 

 capacity of the vessel by multiplying the weight of 

 the vessel by the specific heat of the substance of which 

 it is made. Thus, suppose that the beaker containing 

 the cold water in Experiment 53 weighs 25 g. Its 

 thermal capacity will be 25 x 0*188 = 4- 7 calories. That 

 is to say, to raise its temperature by a given amount 

 will require the supply of as much heat as is required 

 to raise the temperature of 4'7 g. of water by the same 

 amount. Looked at from this point of view, the 

 thermal capacity of the vessel is called its " water- 

 equivalent ". 



Example 2. What will be the temperature attained 

 by the water in Experiment 53, if 100 g. of iron at 

 60 C. be used, and the water, at 15 C., be contained 

 in a beaker weighing 25 g. ? 



The beaker and cold water will be equivalent to 

 a single quantity of water weighing 100 + 4*7 = 104' 7 g. 

 Our equation will hence become 



100 x (60 - n) x 0-114 = 104*7 x (n - 15), 

 or 116-lw = 2254-5, i.e. n = 19'4 C. 



Example 3. Determine the specific heat of zinc 

 from the following data : 



Weight of zinc 300 g. 



beaker 56 g. 



and cold water ... 515 g. 



Temperature of zinc . . . . . . 100 



