Question

The efficiency of heat engine is 0.3. On increasing the temperature of source by 80°C, the efficiency of heat engine increases to 0.4. Find the initial temperature of source and sink in which heat engine is working.

Solution

Let T₁ and T₂ be the initial temperatures of source and sink when the efficiency of the heat engine is 0.3.
Therefore,

0.3 equals 1 minus straight T subscript 2 over straight T subscript 1

rightwards double arrow

straight T subscript 2 space equals space 0.7 straight T subscript 1

...(1)
On increasing the temperature of the source by 80°C, the efficiency becomes 0.4.
Therefore,

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rightwards double arrow

straight T subscript 2 space equals space 0.6 left parenthesis straight T subscript 1 plus 80 right parenthesis

...(2)
From (1) and (2), we get

0.7 straight T subscript 1 space equals space 0.6 left parenthesis straight T subscript 1 plus 80 right parenthesis

rightwards double arrow

0.1 straight T subscript 1 space equals space 48

rightwards double arrow

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, is the temperature of the source.

and

straight T subscript 2 space equals space 0.7 straight T subscript 1 space space space space space space space equals space 0.7 cross times 480 space space space space space space space equals 336 straight K space equals space 63 degree straight C

T₂ is the temperature of the sink.

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