`eta=1-(T_2)/(T_1)=(T_1-T_2)/(T_1)`
When `T_1` is increased by `100^@C`
`eta_1=((1500+100)-(500))/((1500+100+273))=(1100)/(1873)=0.59` i.e., `59%`
When `T_2` is decreased by `100^@C`
`eta_2=(1500-(500-100))/(1500+273)=(1100)/(1773)=0.62` i.e., `62%`
Decreasing `T_2` results in greater improvement in the efficiency.