ALIMENTATION AND THE EXPENDITURE 153 



Calories for a subject of 76 kilogrammes, and -~ x 65 = 1,488 



Calories for a normal adult of 65 kilogrammes in 24 hours. Thus, 

 in complete repose, the muscles being in a state of relaxation, 



the expenditure per kilogramme of the weight of the bod}' is 



1 740 7 



- == 22-89 Calories instead of 32-56 Calories, that is, about -r^. 

 7o 10 



The expenditure of the night (from 7 p.m. to 7 a.m.) is less than 

 that of the day (from 7 a.m. to 7 p.m.), being 43 and 57 hundredths 

 of the total expenditure respectively (fig. 132). In relation to the 

 volume of oxygen the total energy corresponds to 



O OQ7 -y 1 -42Q 



*" * - - 4-97 Calories per litre (0C and 760 millinetres) . 



OoU 



All the heat of interorganic combustion is dissipated externally 

 and is found there as heat, since it has not been used in performing 

 work. The expired air, entering the lungs at 20C and leaving at 

 37C contains this increased heat. As the " specific heat " of a 

 substance is the quantity of heat that is 

 required to raise one gramme of it 1C, we 

 can calculate that portion of the expendi- 

 ture in calories ; the specific heat of the 

 air being 0-2374 calories. In the same way 

 energy is dissipated through the urine and 

 the faeces whose specific heats are taken as 

 1 calorie and 0-9 cal. The water vapor 

 from the lungs and the skin dissipates a 

 considerable fraction of the energy. This 

 is calculated, by finding its heat of vapor- ^ 



isation, which is the quantity of heat needed 



for a gramme to pass into the state of vapour without changing 

 temperature. 



Regnault found for water the well-known relation q = 606-5 

 0-695 X t, t being the temperature of vaporisation, and q the 

 heat evolved expressed in small calories. 



The expired gases carry away vapour which is formed at about 

 37C, the temperature of "the lungs ; hence: q = 606-5 0-695 

 X 37 = 580-8" small calories. 



This vaporised water enters the body at a temperature t' and 

 rises to 37 C, hence the increase of temperature is 37 t', 

 (37 t') calories per gramme are absorbed (the specific heat of 

 the water being unity) . Hence : 



q'= (606-5 0-695 X 37) -f (37 t') =606-5 37(0-695 1 +/') 

 or : 



q' = 606-5 + 37 X 0-305 t'. 



