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The temperature of the sink of a carnot ...

The temperature of the sink of a carnot engine is 300 K and its efficiency is 40 %. Find the decrease in temperature of the sink required to increase the efficiency of the engine to 50% keeping temperature of the source to be constant.

Text Solution

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`eta=1-T_2/T_1`
`therefore 0.4=1-300/T_1`
`therefore 300/T_1=0.6`
`T_1`=?
`T_2`=300 K
`eta`=40% =0.4
`T_1=300/0.6`=500 K….(1)
`eta.=1-(T_2.)/(T_1.)`
`0.5=1-(T_2.)/500`
`(T_2.)/500=0.5`
`T._1`=500 K
`T._2`= ?
`eta`=50% = 0.5
`therefore T._2`=250 K
`therefore` Decrease in temperature of sink
`=T_2-T._2`=300-250=50 K
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