Reversible heat engine converts one sixth of heat input into work. When the temperature of the sink is reduced by 62 C, its efficiency is doubled. Find the temperature of the source and the sink.
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Answer: Temperature of source= 1737 C
Temperature of sink= 1402 C
Explanation:
Let the temperature of the source be T and but the temperature of the sink T'.
The efficiency of the heat engine is given by:
η = 1 - T'/T = W/H
where W is the work done and H is the heat input
It is given that, W = (1/6) H
⇒η = 1 - T'/T = W/H = 1/6
⇒T' = (5/6)T
When, temperature of the sink is reduced by 62 C=335 K, the efficiency is doubled.
⇒2η =(1/3) = 1- (T'-62)/T
⇒(T'-335)/T = (2/3)
Substitute, T' = (5/6)T in the above equation:
(5/6)T -335 = (2/3) T ⇒(1/6)T = 335 ⇒T = 2010 K = 1737 C
T' = (5/6) ×2010 K = 1675 K = 1402 C
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