Question

The efficiency of carranots cycle is 1/6. By lowering the temperature of sink by 65k, it increases to 1/3. calculate the initial and final temperatures of the sink.

Answer 1

The efficiency of a Carnot's engine working between temperatures T1 and T2 is given byη=1−T2T1According to first case,η=1/6=1−T2T1−−−−−−−−−−−(i)Second case,The temperature is reduced by 650CSo,η=1/3=1−T2−65T1−−−−−−−−(ii)Solving equation (i) and (ii),T1=390K=390−273=1170CAnd T2=325K=325−273=52°C

Or

The efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .


condition 1 :- when η ( efficiency of engine ) = 1/6
then, 1/6 = 1 - T₂/T₁ ---------(1)

condition 2 : when temperature of sink is reduced by 65K then, η(efficiency of engine ) = 1/3
e.g, 1/3 = 1 - (T₂ - 65)/T₁ -----------(2)

solve equations (1) and (2)
2(1 - T₂/T₁) = 1 - (T₂ - 65)/T₁
2 - 1 = 2T₂/T₁ - (T₂ - 65)/T₁ = (T₂ + 65)/T₁
T₁ = T₂ + 65
Put it in equation (1)
1/6 = 1 - T₂/(T₂ + 65)
1/6 = (65)/(T₂ + 65)
T₂ + 65 = 390 ⇒ T₂ = 325K or 52°C
now T₁ = 390K or 117°C