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A carnot engine whose efficiency is 40 %...

A carnot engine whose efficiency is 40 % recives heat at 500 K . If the efficiency is to be 50 %. The source temperature for the same exhaust temperature is

A

900 K

B

600 K

C

700 K

D

800 K

Text Solution

Verified by Experts

The correct Answer is:
B
Efficiency of Carnot engine, `b= 1- T_(2)/T_(1)`
Where ` T_(1) and T_(2) ` be the temperatures of source and sink respectively .
` T_(2)/T_(1) = 1 -n=1 - 40/100 =60/100 = 3/5 `
` T_(2) = 3/5 T_(1) = 3/5 xx 500 K = 300 K `
Let ` T_(1)` be the temperature of the source for the same sink temperature.
` therefore T_(2)/T_(1) = 1-n =1 - 50/100 = 1/2`
` T_(1) = 2T_(2) = 2xx 300 K = 600 K `
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