STATICS 125 



investigated in a similar manner. The latent heat of water at 

 40 C. is 580 units, i.e. to evaporate i Ib. of water at that 

 temperature, as much heat is required as would raise the tem- 

 perature of 580 Ibs. of water i C. 0-58 kt. Therefore 



where M is the mass (in pounds) of the water evaporated, and 

 H is the heat required. 



Example. An ox consumed, daily, on the average 



26 Ibs. hay containing 15% water = 3-9 Ibs. water. 

 And 6J gallons of water . . . = 65 



Total water consumed . 68-9 

 The average daily droppings were 



48 Ibs. dung containing 79*7% of water = 38-26 Ibs. water. 

 14 Ibs. urine 88-0% =12-32 



Total water excreted in liquid form . 50*58 



The difference, [68-9 50-6=] 18-3 Ibs., is the average 

 daily amount of water evaporated. The amount of heat 

 consumed in this way is therefore 



18-3 X 0-58 = 10-614 kt. 



Assuming that the available energy of the hay was 1*83 kt. 

 per Ib., 5-8 Ibs. of the food would be consumed in this way. 



At higher temperatures, e.g. in very warm weather, a greater 

 amount of water is evaporated and a larger quantity of food is 

 used for the purpose. 



The available energy of the food required for the main- 

 tenance of an ox (1000 Ibs. live weight) in stationary condition 

 is about 35 kt. per day. Allowing 2 kt. for warming the 

 ingesta and 10 kt. for evaporation of water, there remains 

 23 kt. unaccounted for. A small portion of this is used for 

 warming the respired air, and the remainder over 60 per 

 cent, of the whole l is lost by radiation from the skin. 



Theoretically, for any given animal, the loss of heat by 



1 N.B. This is merely a hypothetical case. The amount of heal 

 actually radiated from the skin is variable but always large. 



