Two identical finite bodies of constant heat capacity at temperatures T₁ and T₂ are available to do work in a heat engine. The final temperature T_f reached by the bodies on delivery of maximum work is

Explanation:

Assume T₁ > T₂

Q₁ = c_p (T₁ - T_F)

Q₂ = c_p (T_F – T₂)

∴ W = Q₁ – Q₂ = c_p (T₁ + T₂ – 2T_F)

For ‘w’ to be max, T_F to be minimum

\({\rm{\Delta }}{S_1} = \mathop \smallint \limits_{{T_1}}^{{T_F}} {C_p}\frac{{dT}}{T} = {C_p}\;ln\frac{{{T_F}}}{{{T_1}}}\)

\({\rm{\Delta }}{S_2} = \mathop \smallint \limits_{{T_2}}^{{T_F}} {C_p}\frac{{dT}}{T} = {C_p}\;ln\frac{{{T_F}}}{{{T_2}}}\)

As,

\({\left( {{\rm{\Delta }}S} \right)_{univ}} \ge 0\;\;\;\therefore {C_p}\;ln\frac{{{T_F}}}{{{T_1}}} + {C_p}\;ln\frac{{{T_F}}}{{{T_2}}} \ge 0\)

\(\therefore {C_p}\;ln\left[ {\frac{{T_F^2}}{{{T_1}{T_2}}}} \right] \ge 0\)

Now,

For T_F to be minimum

\(ln\frac{{T_F^2}}{{{T_1}{T_2}}} = 0\;\;\;\therefore ln\frac{{T_F^2}}{{{T_1}{T_2}}} = ln\;1\)

\(\therefore {T_F} = \sqrt {{T_1}{T_2}} \)