Q: A Carnot engine operating between temperatures T_{1} and T_{2} has efficiency 1/6. When T_{2} is lowered by 62 K, its efficiency increases to 1/3. Then T_{1} and T_{2} are, respectively .

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Sol: $\large \eta_1 = 1-\frac{T_2}{T_1} $

$\large \frac{1}{6} = 1-\frac{T_2}{T_1} $

$\large \frac{T_2}{T_1} = \frac{5}{6} $ … (i)

$\large \eta_2 = 1-\frac{T_2 – 62}{T_1} $

$\large \frac{1}{3} = 1-\frac{T_2 -62}{T_1} $ …(ii)

On solving Equation (i) and (ii)

T_{1} = 372 K and T_{2} = 310 K