CAEBON DIOXIDE AT DIFFERENT TEMPERATURES. 213 



Using (5) 



e m ie H = {l + tAP + Yl[l( m + n)-l-] + 2(n-l) + (n-l)(m-l)y}. 



Thus, for a given value of 6 a , the fraction of 6 m which depends upon the flow is given 

 to the first order by 



{Al 2 +6yl[^(m + n)-l] + 6(n-l)(m-l) r }/6(2n-l) .... (6). 



Now the heat loss between x = and a 1 = I is approximately \lfpQ, and that between 

 x = I and x = n is approximately (n l}fp6. Hence, if k is the quantity defined 



on p. 200, 



HO = 



A/ = 1-y-aBl = hl/(n-$l) JSQ-CV aZ/JSQ. 

 Since CV / = JSQ0 approximately 



A/ = 



Substituting in (<>) and omitting the term ad, which does not involve the flow, we 

 obtain for the fraction of the heat loss which involves the flow 



== (///JSQ} {l a +6l[(m+n)-f]+6(n-l)(m-l)}/3(2n-l) a . 



For the calorimeter employed we may take / = 4, 111 = H, n = 12, so that 



The heat loss in the case of the experiments at steam temperature formed about 

 8 per cent, of the energy supplied in the large flow, and consequently it may be 

 thought that it is not sufficient to work only to the first order in the above investiga- 

 tion. It must be remembered, however, that all we need in order to correct the results 

 calculated on the assumptions of the equations on pp. 200 and 201 is the percentage 

 variation of the heat loss in the two flows, and since this is only a small fraction of 

 the heat loss itself, it is sufficient to calculate it from the first order terms in the heat 

 loss which involve the flow. 



If Q, is the quick rate of flow, and Q 3 the slow rate of flow, the heat loss is greater 

 in the slow flow than in the quick flow by (0'3) h 2 (Q t Q s )/JSQiQg. It is easy to see 

 that the value of the specific heat calculated from the equations on pp. 200 and 201 must 

 be increased by (0'3) h 2 . lOO/JSQ^ JSQ 2 per cent, of its value. In order to reduce 

 the value of CE/S0 (see p. 200) for one experiment of any set to the value which it 

 would have had if the heat loss per degree excess temperature had been the same in 

 that experiment as in the experiment performed with the largest flow, we must 

 decrease it by (0'3) h 2 (Qj Qa)/JSQiQ ? , where Q 2 is the flow for the experiment in 



