Carnot engine has an efficiency of (1)/(7) . When the temperature of the sink is reduced by 60 K, its efficiency is doubled. The temperature of the source is

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 147^ ^ C

The efficiency of a Carnot engine is given by

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

Given

\eta = \dfrac{1}{7}

\dfrac{1}{7} = 1 - \dfrac{T_2}{T_1} \Rightarrow \dfrac{T_2}{T_1} = \dfrac{6}{7}

When the sink temperature is reduced by

60

K, the efficiency is doubled:

\dfrac{2}{7} = 1 - \dfrac{T_2 - 60}{T_1} \Rightarrow \dfrac{T_2 - 60}{T_1} = \dfrac{5}{7}

Subtracting the second equation from the first:

\dfrac{T_2}{T_1} - \dfrac{T_2 - 60}{T_1} = \dfrac{6}{7} - \dfrac{5}{7}

\dfrac{60}{T_1} = \dfrac{1}{7}

T_1 = 420

K

Converting to Celsius:

T_1 = 420 - 273 = 147^{}^{\circ}\text{C}

Answer:

147^{}^{\circ}\text{C}

Carnot engine has an efficiency of (1)/(7) . When the temperature...

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