A carnot engine has efficiency `1//5` . Efficiency becomes `1//3` when temperature of sink is decreased by 50 K What is the temperature of sink ?

Answer

`300`

Explanation

The efficiency of Carnot engine
`eta=("Work output")/("Heat input")=W/(Q_(H))`
or `eta=(Q_(H)-Q_(L))/(Q_(H))=1-(Q_(L))/(Q_(H))`
Also we can show than
`(Q_(L))/(Q_(H))=(T_(L))/(T_(H))`
`:.eta=1-(T_(L))/(T_(H))`
where `T_(L)` is temperature of sink and `T_(H)` is temperature of hot reservoir.
`1/5=1-(T_(L))/(T_(H))`............i
and `1/3=1-(T_(L)-50)/(T_(H))`..........ii
From eq i
`(T_(L))/(T_(H))=4/5`
`impliesT_(H)=5/4T_(L)`
Substituing vaule of `T_(H)` is equ ii we get
`1/3=1-(T_(L)-50)/(5/4T_(L))`
or `(4(T_(L)-50))/(5T_(L))=2/3`
or `T_(L)-50=2/3xx5/4T_(L)`
or `T_(L)-5/6T_(L)=50`
`:.T_(L)=50xx6=300K`