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the temperature of the sink on the carnot engine is 27 degree if efficiency is 25% temp of source is


sajidasajju507 27th Jul, 2021
Answer (1)
ADITYA KUMAR Student Expert 27th Jul, 2021

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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