MR. MACQUORN RANKINE ON THERMO-DYNAMICS. 
161 
where the curve sought, AM, intersects the latter isothermal line ; K L the apparent 
specific heat of the liquid ; — then making the proper substitutions of the symbols of 
temperature for those of heat, and observing that the operation 
is in this case equivalent to multiplication by V B — v, we have 
A<J>= S r(V B - ! ,)(fo.- 
(62.) 
being an equation between two expressions for the difference between the thermo- 
dynamic functions <P for the curve AB, and for that which passes through a . 
If the specific heat of the liquid is approximately constant, this equation becomes 
fj P IT y. 
AO= ^( VB “^)( f °r r=:7 2 ) = K i' h yP- l °S^r K ( 63 -) 
(40.) Corollary . — Problem. The same data being given as in the preceding problem, 
and the expansion of the liquid by heat neglected , a mass of liquid, having been raised 
from the absolute temperature r 2 to the absolute temperature r 15 is supposed to be allowed 
to evaporate partially, under pressure, without receiving or emitting heat, until its tem- 
perature falls again to r. 2 , at which temperature it is liquefied under constant pressure 
by refrigeration : it is required to find the power developed. 
(Solution.) The power developed is represented by the area of the three-sided dia- 
gram of energy in fig. 21, ABa ; that is to say, by 
which, if K l is nearly constant, becomes 
(64.) 
K L £ 1 hyp.log^ = ^.^r==K L |(r 1 -^)-(r 2 -*).(l+ hyp.log^)|. . (65.) 
(41.) Numerical Example. 
Let one pound avoirdupois of water be raised, in the liquid state, from T 2 = 40° 
Centigrade, to T,= 140° Centigrade. Then 
r 1 -*=T 1 +T 0 = 1 40° + 272i°=412i° Centigrade. 
r a -*=T a +T 0 = 40 o +272i°=312i° Centigrade. 
The mean apparent specific heat of liquid water between those temperatures is 
K l =K w (or Joule’s equivalent) x T006=1398 feet per Centigrade degree ; 
consequently the heat expended is equivalent to 139,800 foot-pounds. 
MDCCCLIV. Y 
