556 



JOHN R. MURLIN 



j 



63 CO 2 weighing (63 X 44 =) 2.772 gm. For oxygen the thermal quo- 

 ient would be (7.810-^2.464=) 3.19 Cal. per gram and for CO 2 

 (7.810 -r- 2.772 =) 2.82 Cal. per gm. Or, on the basis of volume at 

 and 760, 







4.54 Cal. per liter of O 2 

 and 5.44 Cal. per liter of CO 2 



For fat the thermal quotient may be calculated from the following 

 equation: C 57 H 101 O 6 + 80.O 2 = 57 CO 2 + 52 H 2 O 

 Triolein 



From this it follows that 0.884 gm. of this particular fat yielding 

 8.423 Cal. would require 80 molecules of O 2 .weighing (80 x 32 =) 2.560 

 gms. and 57 molecules of CO 2 weighing (57x44=) 2.508 gms. One 

 gram of O 2 therefore has a heat value of (8.423 -=- 2.560 =) 3.29 Cal. 

 and one gram of CO 2 (8.423 -=- 2.508 =) 3.35 Cal. or, on the basis of 

 volume at 0" and 760, 



4.70 Cal. per liter of O 2 . 

 and 6.58 Cal. per liter of CO 2 



For carbohydrate the equation is: C C H 10 O 5 + 6 O 2 = 6 CO 2 + 

 5H 2 O and the thermal quotients are : 5.09 Cal. per liter of O 2 



and 5.09 Cal. per liter of CO 2 

 The results may be summarized as in the table below. 



TABLE 2 

 THERMAL QUOTIENTS ( LEFEVKE ( g ) ) 



To estimate the mean thermal quotient for a mixed diet the* method is a 

 simple one. For example, take the mean food consumption of the average 

 soldier in the training camps (p. 554) namely, 122 gm. protein, 123 gm. 

 fat and 485 gm. carbohydrate. The amount of oxygen required for the 

 combustion of these quantities of the several foodstuffs would be : 



122 gm. Protein x 1.524 = 185.9 gm. O 2 



123 gm. Fat x 2.896 = 356.2 " " 

 485 gm. C.H. x 1.185 = 574.7 " " 



Total 1116.8 " " 



