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Q.

The efficiency of Carnot engine is $50 %$ and temperature of sink is $500 K$ . If temperature of source is kept constant and its efficiency raised to $60 %$ , then the required temperature of the sink will be :-

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a

$500 K$

b

$600 K$

c

$100 K$

d

$400 K$

answer is C.

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Detailed Solution

$η_{1} = 50 % \Rightarrow T_{2} = 500 K η_{2} = 60 % \Rightarrow T_{2} = ? η = (1 - \frac{T_{2}}{T_{1}}) \times 100 % \Rightarrow \frac{50}{100} = 1 - \frac{500}{T_{1}} \Rightarrow T_{1} = 1000 K . . . . . . (i) f o r η = 60 % \frac{60}{100} = 1 - \frac{T_{2}}{1000} \Rightarrow T_{2} = 400 K$

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