An ideal heat engine is working between temperature T1 and T2 has efficiency η . If both the temperatures are raised by 100oK each, the new efficiency will be
A
η
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B
less than η
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C
more thanη
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D
cannot be perdicted
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Solution
The correct option is B less than η The efficiency of a system is defined as, η=T2−T1T2
Now, if both the temperatures are increased, η′=T2−T1T2+100