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The efficiency of a Carnot's engine at a...

The efficiency of a Carnot's engine at a particular source and sink temperature is `(1)/(2)` .When the sink temperature is reduced by `100^(@)C`, the engine efficiency, becomes `(2)/(3)`. Find the source temperature.

Text Solution

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`1-T_2/T_1 = 1/2 …(i)` (`T_2` : sink temperature `T_1` : Source Temperature)
and `1-((T_2-100))/T_1=2/3 …(ii)`
or, `T_2/T_1=1/2 ` from (i)
and `((T_2-100))/T_1 = 1/3` from (ii)
Dividing these two we get
`T_2/(T_2-100)=3/2`
or, `T_2 = 300K`
So, `T_1 = 600K`
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