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A Carnot engine whose sink is at 300 K h...

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?

A

`380 K`

B

`275 K`

C

`325 K`

D

`250 K`

Text Solution

Verified by Experts

The correct Answer is:
d
Efficiency of a Carnot engine, `eta = 1 - T_(2)/T_(1)`
or, `T_(2)/T_(1) = 1 - eta = 1 - 40/100 = 3/5`
` :. T_(1) = 5/3 xx T_(2) = 5/3 xx 300 = 500 K`.
Increase in efficiency = `50% " of " 40% = 20%`
New efficiency, `eta. = 40% + 20% = 60%`
` :. T_(2)/T_(1) = 1 - 60/100 = 2/5`
` rArr T_(1). = 5/2 xx T_(2) = 5/2 xx 300 = 750 K`
` :. ` Increse in temperature of source ` = T_(1). - T_(1)`
` = 750 - 500 = 250 K`
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