172 Domestic Science 



obtained in an actual determination by the above 

 method as follows : 



Weight of beaker . . . . . . . . 84' 6 g. 



and cold water . . . . 341'8 g. 



,, water at end of expt. .. 35T6g. 



Temperature of water at beginning of expt. . . 5' 5 



end .. 26-9 



If L denote the latent heat, we have 

 Heat supplied by steam in condensing to water 

 = (351-6 - 341-8) L cal. 



Heat given out by water formed from the steam in 

 cooling to 26-9 



= (351-6 - 341-8) x (100 - 26'9) cal. 

 Heat received by the cold water and beaker 



= (341-.8 - 84-6 + 15*8) x (26'9 -- 5'5) cal. 



The last quantity of heat is equal to the sum of the 

 other two, i.e. 



273 x 21-4 = 9-8.L + 9'8 x 73'1. 

 Solving this equation in L, we have 



9-8L = 5125-8, or L = 523. 



The results obtained are always somewhat low, since 

 it is practically impossible to assure that only " dry " 

 steam (i.e. steam containing no condensation water 

 whatever) shall pass into the cold water. 



110. From the working out of the results given 

 in the last two experiments, it will be fairly easy to 

 devise methods of solving problems in which latent 

 heats are involved. 



Instead of the glass beakers used in the experiments 

 on specific and latent heats, specially made copper 



