In: Physics
An ideal heat engine takes in heat from a reservoir at 330 °C and has an efficiency of 31%. If the exhaust temperature does not vary and the efficiency is increased to 42%, what would be the increase in the temperature of the hot reservoir?
Solution :-
Case I:
T1 = 330 °c
= 603 K
Efficiency = (T1 - T2) / T1
0.31 = ( 603 - T2) / 603
186.93 = 603 - T2
T2 = 416 K
Case II:
Temeprature of reservoir = T1'
Efficiency = (T1' - T2 ) / T1'
0.42 = ( T1' - 416 ) / T1'
0.42T1' = T1' - 416
T1' ( 1 - 0.42) = 416
T1' = 416 / 0.58
= 717.24 K
= (717.24- 273) deg c
= 444.24 degc
Increase in the temperature of hot reservoir = T1' - T1
= 444.24 - 330
= 111.24 deg