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+1 vote
13.8k views
in Physics by (51.2k points)
closed by

A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be

(1) 360 K 

(2) 1000 K 

(3) 900 K 

(4) 300 K

2 Answers

+2 votes
by (50.5k points)
selected by
 
Best answer

Correct option is (2) 1000 K 

Now efficiency is increased to 70% and T2 = 300 K, Let temp of source T1 = T

+1 vote
by (20 points)
The new temperature of the source will be 1000k

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