Page:Reflections on the Motive Power of Heat.djvu/184

160 the cycle of operations described above, $$\tfrac{H}{dq}$$ times, we have 27. If the amplitudes of the operations had been finite, so as to give rise to an absorption of H units of heat during the first operation, and a lowering of temperature from S to T during the second, the amount of work obtained would have been found to be expressed by means of a double definite integral thus: {{MathForm2|(4)|$$\left. \begin{array}{ll} \qquad M = \int_{0}^{H} dq \int_{T}^{S} dt \cdot \frac{E p_{_0} v_{_0}}{v dq/dv} ,\\ or\\ \qquad M = E p_{_0} v_{_0} \int_{0}^{H} \int_{T}^{S}\frac{1}{v}\frac{dv}{dq}\, \cdot {dtdq}; \end{array} \right\} $$}}

this second form being sometimes more convenient.