• # question_answer A carnot's reversible engine converts $\frac{1}{6}th$ of heat input into work. When the temperature of sink is reduced 62 K, the efficiency of carnot's cycle, becomes $\frac{1}{3}.$ Calculate temperature of source and sink A) 372K, 310K        B) 772K, 312K C) 672K, 610K D) None of these

 [a] $\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}$ where ${{T}_{1}}\And {{T}_{2}}$ are source & sink temperatures respectively. $\frac{1}{6}=1-\frac{{{T}_{2}}}{{{T}_{1}}}$               (1) Modified case, $\frac{1}{3}=1-\frac{{{T}_{2}}-62}{{{T}_{1}}}$         ..(2) Solving (1) and (2), we get ${{T}_{1}}=372K,\,{{T}_{2}}=310\,K$