Latent Heat of Water and Ice. 31 



they now arc, it was supposed that the heat became hidden 

 or latent but that it was heat still. 



42. Measuring the Energy Required to Melt Ice. This 

 may be determined approximately by taking equal weights 

 of water at 212 F and of ice at 32 F., putting the two 

 together and noting the temperature at the moment the ice 

 is all melted. When this has been done it will be found 

 that the combined water has a temperature of about 51 F. 



If, however, equal weights of water at 32 and 212 are 

 mixed there will be found a temperature of 



212 + 32 



one volume of water having lost as much as the other 

 gained. 



In the first case, however, the water lost 



212 51 = 161 



while the ice gained only 



51-32 = 19. 



There was therefore in this case an apparent loss of 

 101 19 = 142 



If a pound of water and of ice had been taken for these ex- 

 periments it is plain from (38) that the 142 would also 

 represent 142 heat units. 



43. Measuring the Energy Required to Evaporate Water. 

 If a pound of steam at 212 F. be condensed within 5.37 

 pounds of water at 32 F. there will result 6.37 pounds of 

 water having a temperature very close to 212 F. The 

 one pound of steam has therefore raised the temperature of 

 5.37 pounds of water through 



2 12 -32 == 



