564: / ; JOHN K. MUKLIISr 



long experiments Kauffmann undertook to find an exact heat equivalent 

 for any particular short period by what he called successive thermal quo- 

 tients. This means only that he partitioned the oxygen to the several 

 organic foodstuffs and multiplied hy their respective thermal quotients. 

 For example in an experiment on a dog subjected to a prolonged fast he 

 found that the animal had absorbed in 1 hour 5.992 liters of 2 , had 

 given off 4.494 1. of CO 2 and eliminated 0.1983 gm. N" in the urine. The 

 E. Q. was 0.75. The nitrogen corresponded to (0.1983 X 6.25 =) 1.239 

 gm. protein burned, which in turn required 1.72 gm. of O 2 to oxidize it to 

 the stage of urea (page 555). Subtracting this from the total oxygen 

 (5.992 1. = 8.57 gm.) there remained 6.85 gm. for combustion of fat. The 

 heat production was found as follows : 



1.72 gm. 2 X 3.19 = 5.486 Cal. 

 6.85 " " X 3.29 = 22.536 " 

 Total 28.022 " 



Applied to the human subject in good nutritive condition and sub- 

 sisting on a mixed diet the method would be a little more complicated. 

 Thus Arthus reports the metabolism of a man for 24 hours: 



O 2 absorbed = 496 1. or 709 gm. 



CO 2 eliminated = 463 1. or 912 gm. 



N" in urine 17.35 gm. = 108.44 gm. protein 



The protein would require the absorption of 151 gm. O 2 and elimination 

 of 180 gm. CO 2 . 



709 151 558 gm. O 2 or 390 1. 

 912 180 = 732 " CO 2 or 371 1. 



The remainder represents the metabolism of carbohydrate and fat. 



Let x be the volume of O 2 for combustion of fat and y the volume of 

 CO 2 resulting. Let z represent the volume of O 2 and CO 2 for combus- 

 tion of carbohydrate. 



Then y x= 0.70 

 x + z =390 1. 

 y + z = 371 1. 

 From which x = 63.33 liters O 2 

 y= : 44.33 " CO 2 

 z = 326.33 " O 2 and C0 2 



The weights of a liter of O 2 at 760 mm. Hg and being 1.43 grams* 

 the apportionment of O 2 would be as follows : 



For fat (63.33 x 1.43 =) 90.56 gm, 

 " carbohydrate 467.12 " 



" protein 151.0 " 



