Question

A stone with heat capacity C = 1.2 J/K is left outside on a cold day to reach a temperature of 273.15 K. The stone is then brought inside where the air temperature is 293.15 K. The stone is used as the cold side of a reversible engine. (The air is the hot side.) What is the maximum work that can be accomplished ?

Answer 1

To solve the problem it is necessary to apply the concepts related to Helmholtz free energy. By definition in a thermodynamic system the Helmholtz energy is defined as

$\Delta F = \Delta U - T\Delta S$

Where,

$\Delta U$ is the internal energy equivalent to

$\Delta U = C \Delta T$

And $\Delta S$ means the change in entropy represented as

$\Delta S = C ln \frac{T_2}{T_1}$

Note: C means heat capacity.

Replacing in the general equation we have to

$\Delta F = C \Delta T - T C ln \frac{T_2}{T_1}$

The work done of a thermodynamic system is related by Helmholtz free energy as,

$W = - \Delta F$

$W = -(C \Delta T - T C ln \frac{T_2}{T_1})$

$W = T C ln \frac{\T_2}{T_1}-C \Delta T$

Replacing with our values we have,

$W = (293.15)(1.2)ln(\frac{293.15}{273.15})-(1.2)(20)$

$W = 0.858 J$

Therefore the maximum work that can be accomplished is 0.858J