588 AN AMERICAN TEXT-BOOK OF PHYSIOLOGY. 



V P 

 obtained by the following formula : V = , where V is 



760 (1 + 0.003665 t) 



the required volume at C. and 760 mm. barometric pressure, V the ob- 

 served volume, P the observed pressure, and t the observed mean temperature : 

 760 (1 + 0.003665) is conveniently obtained from standard tables. The errors 

 incident to changes in barometric pressure and in aqueous tension are so slight 

 that they are not usually taken into consideration. Assuming that the quan- 

 tity of air supplied amounted to 6000 liters, and that the mean temperature 

 of the air was 20, the corrected volume would be, omitting barometric 



V 6000 



pressure and aqueous tension, V = - = 5590 liters 



(1 + 0.0036656 t) 1,0733 



at C. One litre of dry air at C. weighs 0.001293 kilogram ; therefore, 

 5590 liters X 0.001293 = 7.228 kilograms. If we assume that the air during 

 its passage through the calorimeter had its temperature increased 3, and the 

 specific heat of air is 0.2377, the quantity of heat imparted to the air must 

 have been 7.228 X 3 X 0.2377 = 5.152 kilogramdegrees. 



The next estimate is of the quantity of heat lost in the evaporation of 

 water. This is determined by finding the difference between the quantities 

 of water in the samples of the air passing into and from the calorimeter, and 

 estimating from these results the amount of moisture imparted to the total air 

 leaving the chamber. Assuming that 10 grams of water were thus evaporated, 

 since each gram requires about 582 calories or 0.582 kilogramdegree, the quan- 

 tity of heat evolved would be equal to 10 X 0.582 = 5.82 kilogramdegrees. 



The total quantity of heat dissipated would therefore be the sum of the 

 quantities given to the calorimeter, to the air, and to the water evaporated : 



Given to the calorimeter 54,000 kilogramdegrees. 



Given to the air 5,152 " 



Lost in evaporating water 5,820 



Total heat-dissipation 64,972 



The quantity of heat produced is determined by adding to or subtracting 

 from the quantity dissipated the amount of heat that may have been gained 

 or lost by the organism. It is obvious that any difference between the 

 quantities of heat dissipated and produced must be represented by an increase 

 or decrease of the mean temperature of the animal. If the animal's tempera- 

 ture remains unchanged, the quantity of heat produced is the same as the 

 quantity lost; if, however, the animal's temperature increases, less heat is 

 dissipated than is produced; if it falls, vice versa. The quantity of heat 

 involved in a change of body-temperature is determined by the product of 

 the change in temperature into the animal's weight and specific heat. Assum- 

 ing that the animal's temperature at the beginning of the experiment was 

 38.95 C. and at the end 39.32 C., the temperature being increased 0.37 C., 

 that the animal's weight was 25 kilograms, and that the animal's specific heat 

 was 0.8, the quantity of heat would be 0.37 X 25 X 0.8 = 7.4 kilogramdegrees. 



