Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
A Carnot engine has efficiency of 50 %. If the temperature of sink is reduced by 40° C, its efficiency increases by 30 %. The temperature of the source will be :
Q. A Carnot engine has efficiency of
50%
. If the temperature of sink is reduced by
4
0
∘
C
, its efficiency increases by
30%
. The temperature of the source will be :
612
112
JEE Main
JEE Main 2022
Thermodynamics
Report Error
A
166.7
K
B
255.1
K
C
266.7
K
D
367.7
K
Solution:
η
=
1
−
T
H
T
L
2
1
=
1
−
T
H
T
L
2
1
(
1
⋅
3
)
=
1
−
(
T
H
T
L
−
40
)
2
1
(
1
⋅
3
)
=
2
1
+
T
H
40
T
H
=
266.7
K