• # question_answer A Carnot engine takes $3\times {{10}^{6}}$cal of heat from a reservoir at $627{}^\circ C$ and gives it to a sink at $27{}^\circ C$. The work done by the engine is A)  $4.2\times {{10}^{6}}\,J$              B)   $8.4\times {{10}^{6}}J$ C)  $16.8\times {{10}^{6}}J$              D)  zero

Given, ${{T}_{1}}=627+273=900\,K$ ${{Q}_{1}}=3\times {{10}^{6}}\,cal$ ${{T}_{1}}=27+273=300\,K$             At,        $\frac{{{Q}_{1}}}{{{T}_{1}}}=\frac{{{Q}_{2}}}{{{T}_{2}}}$ $\Rightarrow$            ${{Q}_{2}}=\frac{{{T}_{2}}}{{{T}_{1}}}\times {{Q}_{1}}$ $=\frac{300}{900}\times 3\times {{10}^{6}}$ $=1\times {{10}^{6}}\,cal$ Work done $={{Q}_{1}}-{{Q}_{2}}$ $=3\times {{10}^{6}}-1\times {{10}^{6}}=2\times {{10}^{6}}\,cal$ $=2\times 4.2\times {{10}^{6}}J=8.4\times {{10}^{6}}J$