Page:Reflections on the Motive Power of Heat.djvu/193

latent heat of a pound of saturated steam, by the equation^[1]

$k={\frac {p}{760}}\cdot {\frac {1+.00366\times 100}{1+.00366\times t\quad }}\cdot \times {.036869}\,.\lambda ,$

where p denotes the pressure in millimetres, and λ the latent heat of a pound of saturated steam; the values of λ, being calculated by the empirical formula^[2]

$\lambda =(606.5+0.305t)-(t+.00002t^{2}+0.0000003t^{3}),$

given by Regnault as representing, between the

1. It appears that the vol. of 1 kilog. must be 1.69076 according to the data here assumed.

   The density of saturated steam at 100° is taken as ${\tfrac {1}{1693.5}}$ of that of water at its maximum. Rankine takes it as ${\tfrac {1}{1696}}$.

2. The part of this expression in the first vinculum (see Regnault, end of ninth memoir) is what is known as "the total heat" of a pound of steam, or the amount of heat necessary to convert a pound of water at 0° into a pound of saturated steam at t°; which, according to "Watt's law" thus approximately verified, would be constant. The second part, which would consist of the single term t, if the specific heat of water were constant for all temperatures, is the number of thermic units necessary to raise the temperature of a pound of water from 0° to t°, and expresses empirically the results of Regnault's experiments on the specific heat of water (see end of the tenth memoir), described in the work already referred to.