the temperature of the sink on the carnot engine is 27 degree if efficiency is 25% temp of source is
Hello
Answer :- Temperature of source = 127 °C
EXPLANATION
We have
Efficiency, η = 25% = 25/100 = 1/ 4
Temperature of the sink T2 = 27 °C = 300 K
let us assume temperature of source be T1.
we know that,
Efficiency, η = [ 1 – (T2 / T1) ]
(1/4) = 1 – [(300) / T1]
Or (3/4) = (300) / T1
Or T1 = 400 K
Or Temp of source = 400 K = 400 - 273 = 127 °C
Thank you


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