Answer to Question #120750 in Molecular Physics | Thermodynamics for Abhishek

Question

A reversible engine converts one-sixth of the heat input into work. If the temperature of the sink is reduced to 335 K, its efficiency is doubled. Calculate the temperatures of the source and the sink.

Answer 1

The efficiency can be calculated as

$\eta = 1 - \dfrac{T_2}{T_1}$ , where $T_1$ is the source temperature and $T_2$ is sink temperature.

Initially $\eta_1 = \dfrac16$ and after reducing the temperature $\eta_2 = 2\eta_1=\dfrac13$ . We may determine the temperature of the source:

$\eta_2 = 1- \dfrac{335\,\mathrm K}{T_1} = \dfrac13, \;\; T_1 = 502.5\,\mathrm{K}.$

Next, we can obtain the initial temperature of sink:

$\eta_1 = 1 - \dfrac{T_2}{502.5\,\mathrm{K}} = \dfrac16, \;\; T_2 = 418.75\,\mathrm{K}.$