A reversible engine has an efficiency of 1/4. If the temperature of the sink is reduced by 58^{} {C}, its efficiency becomes double. Calculate the temperature of the sink. [31 Aug 2021 Shift 1]

Correct Answer is : 174∘C174^{} {C}174∘C

Correct Answer is : 174∘C174^{} {C}174∘C

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Thermodynamics Part 3

Section: Physics

Question : 65 of 100

Marks: +1, -0

\frac{1}{4}

58^{\circ} \mathrm{C}

Solution:

Given, initial efficiency of engine,

\eta_1 = \frac{1}{4}

Let, initial temperature of

\sin k = T_2

Final temperature of

\sin k = T_2 - 58^{\circ} \mathrm{C}

Final efficiency of engine,

\eta_2 = 2 \eta_1 = \frac{2}{4} = \frac{1}{2}

Temperature of source

= T_1

Since,

\eta = 1 - \frac{T_2}{T_1}

\therefore \eta_1 = \frac{1}{4} = 1 - \frac{T_2}{T_1}

\Rightarrow \frac{T_2}{T_1} = 1 - \frac{1}{4}

\Rightarrow \frac{T_2}{T_1} = \frac{3}{4}

\Rightarrow T_1 = \frac{4}{3} T_2

...(i)

and

\eta_2 = \frac{1}{2} = 1 - \frac{T_2 - 58}{T_1}

\Rightarrow \frac{T_2 - 58}{T_1} = 1 - \frac{1}{2}

\Rightarrow \frac{T_2 - 58}{\frac{4}{3} T_2} = \frac{1}{2}

[From Eq. (i)]

\Rightarrow 2 T_2 - 116 = \frac{4}{3} T_2

\Rightarrow \frac{2 T_2}{3} = 116

\Rightarrow T_2 = \frac{116 \times 3}{2} = 58 \times 3 = 174^{\circ} \mathrm{C}

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