170 



Domestic Science 



Original temperature of water . . . . 21 

 Final .. .. 11 



Let the latent heat of fusion of ice be denoted 

 by L. 



Water-equivalent of beaker = 84- 6 x 0-188 = l'8. 

 Heat supplied by water and beaker 



= (224-3 + 15'8) x (21 - 11) cal. 



Fig. 52. 



Heat rendered latent during fusion 

 = (335-6 - 308-9) x Leal. 



Heat needed to raise temperature of water formed 

 by fusion of the ice from to 11 



= (335-6 - 308-9) x 11 cal. 



Equating the first quantity of heat to the other 

 two, we have 26'7 = 2107'3, or L = 79. 



