
MECHANICAL ACTION OF HEAT. 177 
this determination may be considered correct to about jgg5 part. When French 
measures are used in the calculation, the following is the result :— 
v=1 cubic centimetre per gramme, 
@=1033:3 grammes per square centimetre, 
1 : , 
CnM~ 46-78 métres per centigrade degree, 
—153°48 feet aes ie oh oi Hiya ih Alia) 
or 85:27 feet per degree of Fahrenheit. 

The difference, which is of no practical importance in calculating the power 
of the steam-engine, arises in the estimation of the density of liquid water. 
(22.) Unit of weight of steam at saturation, of the elasticity P, and volume V, 
corresponding to the absolute temperature 7,, being cut off from external sources 
of heat, it is now to be investigated what amount of power it will produce in 
expanding to a lower pressure P, and temperature 7,. 
It has already been shewn. at the end of the second section, that if vapour at 
saturation is allowed to expand, it requires a supply of heat from without to main- 
tain it at the temperature of saturation, otherwise a portion of it must be liquefied 
to supply the heat required to expand the rest. Hence, when unity of weight of 
steam at saturation, at the pressure P, and volume V,, expands to a lower pressure 
P, being cut off from external sources of heat, it will not occupy the entire volume 
V corresponding to that pressure, according to Equation (38.), but a less volume 
S=m V, 
where m represents the weight of water remaining in the gaseous state, the por- 
tion 1—m having been liquefied during the expansion of the remainder. The 
expansive action of the steam will therefore be represented by 
fas 2. seine) 
Wa 
The law of variation of the fraction m fiows from the following considera- 
tions :— 
Let 6 m represent the indefinitely small variation of m corresponding to the 
indefinitely small change of temperature d7; L, the latent heat of evaporation of 
unity of weight; K,, as in Equation (30.), the specific heat of vapour at satura- 
tion, which is a negative coefficient varying with the temperature; then we must 
have 
—Lom=mK, 67, or om ES gn 
m L 
in order that the heat produced by the liquefaction of dm may be equal to the 
heat required to expand m. Hence making, according to Equation (30.)— 
K, r= (67+N— OV) 
