A Cannot engine whose efficiency is 40%, receives heat at 500 K. If the efficiency is to be 50%, the source temperature for the same exhaust temperature is

Question

A Cannot engine whose efficiency is 40%, receives heat at 500 K. If the efficiency is to be 50%, the source temperature for the same exhaust temperature is (A) 900 K (B) 600 K (C) 700 K (D) 800 K

Tardigrade Admin · Accepted Answer

Cannot efficiency relation η =(T1-T2/T1) Case I (T1-T2/T1)=0.4 T1-T2=0.4T1 T2=0.6T1 Case II (T1-T2/T1)=0.5 T1=(0.6/0.5)T1=600 K

Q. A Cannot engine whose efficiency is 40%, receives heat at 500 K. If the efficiency is to be 50%, the source temperature for the same exhaust temperature is

900 K

600 K

700 K

800 K

550 K

Cannot efficiency relation $η = \frac{T _{1} - T _{2}}{T _{1}}$
Case I $\frac{T _{1} - T _{2}}{T _{1}} = 0.4$
$T_{1} - T_{2} = 0.4 T_{1}$
$T_{2} = 0.6 T_{1}$
Case II $\frac{T _{1} - T _{2}}{T _{1}} = 0.5$
$T_{1} = \frac{0.6}{0.5} T_{1} = 600 K$