BETWEEN THE FREEZING AND BOILING-POINTS. 
157 
water, Q t and Qo, for the same inflow temperature d 0 , then we have the two 
equations 
4-2 Q, (0 t - e„) S, + (tf, - e 0 ) h = EjC, - 4-2 Q, ( 6\ - 0 O ), 
4'2 Qo (d 2 — d 0 ) So + (do — 9 0 ) h = E 2 C 3 — 4 - 2 Q 2 (do — d 0 ). 
If the electric current is adjusted for the two flows so that d t = do, then 
(d x - d 0 ) h = (0 i — d 0 ) h. 
and Si = So 
S, and hence by direct subtraction and writing t/d = (d 1 —d 0 ) = (do —d 0 ), 
4-2 (Qj - Qo) d0S = (E l C l - 4-2 Qj dd) - (E 2 C 2 - 4'2Q 3 cZd), 
and 
S = 
(EjC! - 4-2 Q, dd) - (E 2 C 3 - 4-2 Q 3 d0) 
4-2 (Q x — Qo) dd ~ ~ ~ ’ 
from which J, or the number of joules per calorie = 4'2 (1 ^ §). By substituting S 
in either difference-equation, h can be obtained. 
The value of J thus obtained will be the mean over the range cl6 through which 
the water is heated, and apply to the mean temperature 
+ 2 (#1 - 0 O ) = T (mean). 
If the variation of the value of J is not linear over this mean temperature, then for 
different values of d 0 and 9 l for the same value of T (mean), the value of J will be 
slightly different. 
Application of the General Difference Equation to Test the Theory of the Method .— 
In order to test the accuracy of the assumptions made in regard to 
(a) The dependence of the heat-loss on the rise of temperature, 
(b) The dependence of the heat-loss on the flow, including the conduction 
correction, 
we will consider the general difference equation. We have as before 
4-2 Q (cW) S + (dd) h = EC — 4-2 Q (dO). 
Dividing through by d9 , the equation is expressed per degree rise, or 
4-2QS + A = 
EC - 4-2 Q dd 
dd 
