The efficiency of Carnot engine is y, when it `eta` hot and cold reservoirs are maintained at temperature `T_(1)` and `T_(2)`, respectively. To increase the efficiency to `1.5 eta`, the increase in temperature `(DeltaT)` of the hot reservoir by keeping the cold one constant at `T_(2)` is 

Options

A. `(T_(1)T_(2))/((1-eta)(1-1.5eta))`
B. `(0.5T_(2)eta)/((1-1.5eta)(1-eta))`
C. `(T_(1))/((1-eta))-(T_(2))/((1-1.5eta))`
D. `((1-eta)(1-1.5eta))/(T_(1)T_(2))`

Answer

B

Explanation

Efficiency of carnot cycle is
`eta=1-(T_(2))/(T_(1))`
`implies" "1-eta=(T_(2))/(T_(1))implies(T_(1))/(T_(2))=(1)/((1-eta))" "…(i)`
In second case,
`1.5eta=1-(T_(2))/(T_(1)+DeltaT)`
or `" "1-1.5eta=(T_(2))/(T_(1)+DeltaT)`
`implies" "(T_(1)+DeltaT)/(T_(2))=(1)/((1-1.5eta))" "...(ii)`
Using Eqs. (i) and (ii), we get
`(Delta T)/(T_(2))=(1)/(1-1.5eta)-(1)/(1-eta)`
`=(1-eta-1+1.5eta)/((1-eta)(1-1.5eta))`
`implies" "DeltaT=(0.5etaT_(2))/((1-eta)(1-1.5eta))`