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Heat Engines

A Carnot engi...

Question

A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency 0.2. When $T_{2}$ is reduced by 60 K, its efficiency increases to 0.4. Then, $T_{1}$ and $T_{2}$ are respectively

200 K, 150 K

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250 K, 200 K

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300 K, 240 K

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300 K, 200 K

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300 K, 150 K

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Solution

The correct option is C 300 K, 240 K
Maximum efficiency of an engine working between temperature $T_{1}$ and $T_{2}$ is given by the fraction of the heat absorbed by an engine which can be converted into work is known as efficiency of the heat engine.
Mathematically,
Efficiency, $η = (T_{1} - T_{2}) / T_{1}$
Subsituting value of $η = 0.2$ ,
we get $0.8 T_{1} = T_{2}$
Now subsituting value of $η = 0.4$ and sink temperature as $T_{2} - 60$ ,we get $0.6 T_{1} + 60 = T_{2}$
Solving both we get $T_{1} = 300 K$ and $T_{2} = 240 K$

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A Carnot engine operating between temperatures T1 and T2 has efficiency 0.2. When T2 is reduced by 60 K, its efficiency increases to 0.4. Then, T1 and T2 are respectively

A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency 0.2. When $T_{2}$ is reduced by 60 K, its efficiency increases to 0.4. Then, $T_{1}$ and $T_{2}$ are respectively