A Carnot engine $(\mathrm{E})$ is working between two temperatures 473 K and 273 K . In a new system two engines - engine $E_1$ works between 473 K to 373 K and engine $E_2$ works between 373 K to 273 K . If $\eta_{12}, \eta_1$ and $\eta_2$ are the efficiencies of the engines $E, E_1$ and $E_2$, respectively, then
Solution
<p>We know, efficiency of a carnot engine, $\eta$ or $e = 1 - {{{T_C}} \over {{T_H}}}$</p>
<p>where, T$_C$ = temperature of cold sink</p>
<p>T$_H$ = temperature of hot source</p>
<p>So, ${\eta _{12}} = 1 - {{273} \over {473}} = {{200} \over {473}} = 0.423$</p>
<p>${\eta _1} = 1 - {{373} \over {473}} = {{100} \over {473}} = 0.211$</p>
<p>${\eta _2} = 1 - {{273} \over {373}} = {{100} \over {373}} = 0.268$</p>
<p>Here, ${\eta _1} + {\eta _2} = 0.211 + 0.268 = 0.479 > 0.423$</p>
<p>So, ${\eta _{12}} < {\eta _1} + {\eta _2}$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.