1.

A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency:

Answer»

380 K
275 K
325 K
250 K

Solution :`eta =1-(T_(2))/(T_(1)) rArr (40)/(100)=1-(300)/(T_(1)) rArr T_(1)=500 K`
New EFFICIENCY.
`eta. =eta +(50)/(100) cdot eta =40% +(50)/(100)xx40% =60%`
`therefore eta.=1-(T_(2))/(T_(1)) rArr (60)/(100)=1-(300)/(T_(1))w`
`rArr T_(1) =750 K`
`therefore` Increase in temperature of SOURCE
`=T_(1).-T_(1)=750-500=250 K`.
`therefore` Correct choice is (d).


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