InterviewSolution
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InterviewSolution
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An engine has an efficiency of 0.25 when temperature of sink is reduced by `58^(@)C` , If its efficency is doubled, then the temperature of the source isA. `150 ^(@)C`B. `222 ^(@)C`C. `242 ^(@)C`D. `232 ^(@)C` |
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Answer» Correct Answer - D Here `eta_(1)=1-(T_(2))/(T_(1))` or 0.25 =1 -`(T_(2))/(T_(1)) rArr (1)/(4) =1-(T_(2))/(T_(1))` `(T_(2))/(t_(1))=1-(1)/(4)=(3)/(4)` According to question, `eta_(2)=2eta_(1)` and `T_(2)=T_(2)-58^(@)C` `therefore 2xx(1)/(4)=1(T_(2)-58^(@)C)/(T_(1)) rArr (1-(1))/(2)=(T_(2)-58^(@)C)/(T_(1))` `(1)/(2)=(T_(2))/(T_(1))-(58^(@))/(t_(1))rArr(3)/(4)-(1)/(2)=(58)/(T_(1)) rArr T_(1)=232^(@)C` |
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