200 MAX KLEIBER 



reading on the galvanometer, G, when i7 is at a given temperature, 

 say 37°C. Since the resistance of the copper wire varies with the 

 temperature, and the resistance of the manganin wire is practically 

 independent of temperature changes, the bridge is balanced only when 

 H is at the temperature for which the instrument is adjusted. Since 

 the resistance of the bridge in equilibrium is a constant, the heat neces- 

 sary to keep the test body at the desired temperature may be calcu- 

 lated from the intensity of the current regulated by the rheostat, R, 

 and read on the ammeter, A. 



10. Indirect Calorimetry 



Animal calorimetry measures the heat loss of animals. When the 

 heat capacity and temperature of a body remain constant, the rate of 

 heat loss is equal to the rate of heat production. In short experi- 

 ments heat production and heat loss may differ significantly. 



Animal heat production is now usually estimated from chemical 

 measurements. This procedure is known as indirect calorimetry. 

 It is based on the law of Hess, mentioned above, and deals only with 

 initial and final conditions, being unaffected by the particular processes 

 known as intermediary metabolism, which characterize the change. 

 A given decrease of carbohydrate, protein, and fat in the animal body 

 and a given consumption of oxygen, accompanied by a corresponding 

 production of excretory products, such as urea and carbon dioxide, 

 indicate a definite heat production. 



The heat of combustion of the major group of nutrients is given 

 in Table IV. The heat of combustion is, as a rule, measured in a 



TABLE IV 

 Caloric Equivalents of Nutrients 



" Calculation for protein: 



Heat of combustion per gram nitrogen in protein 5 . 7 X 6 . 25 = 35 . 6 kcal. 

 Loss in urine: heat of combustion per mole urea = 152 kcal. 



per gram nitrogen in urea = 152/28 = 5.4 kcal. 



Difference'. Catabolizable energy per gram nitrogen in protein = 30.2 kcal. 



Catabolizable energy per gram protein = 30.2/6.25 = 4.8 kcal. 



