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Q.

The efficiency of a carnot engine is 50%. The temperature of the hot reservoir is kept constant. By what amount should the temperature of the cold reservoir be decreased so that efficiency becomes 60%.

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a

1K

b

2K

c

30K

d

83K

answer is A.

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Detailed Solution

0.5 = 1 – TC/TH.And 0.6 = 1 – (TC + x)/TH.From the first equation we get: 5/10 = TC/TH.From the second equation we get: 4/10 = (TC + x)/TH.Now, we see that the value of x must be 1K.
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