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KEAM 2019 Physics and Chemistry Paper
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© examsnet.com
Question : 58
Total: 120
A Carnot engine is operating between a hot body and cold body maintained at temperature
T
1
and
T
2
respectively. Consider the following three cases
Case-I : The temperature of the hot body is changed to
T
1
+ αT and cold body is at
T
2
Case II: The temperature of the hot body is at
T
1
and cold body is changed to
T
2
+ αT
Case –III: The temperature of the hot body is at
T
1
and cold body is changed to
T
2
– αT
The efficiency of the Carnot cycle is highest for case -I
The efficiency of the Carnot cycle is highest for case –II
The efficiency of the Carnot cycle is highest for case –III
The efficiency of case –II is higher than case –III
The efficiency of the Carnot cycle is same for all three cases
Validate
Solution:
We know that efficiency of Carnot engine is given by relation
η
=
T
1
−
T
2
T
1
Hence in
Case I
η
1
=
(
T
1
+
∆
T
)
−
T
2
T
1
+
∆
T
=
(
T
1
−
T
2
)
+
∆
T
T
1
+
∆
T
Case II
η
I
I
=
T
1
−
(
T
2
+
∆
T
)
T
1
Case III
η
I
I
I
=
T
1
−
(
T
2
−
∆
T
)
T
1
=
(
T
1
−
T
2
)
+
∆
T
T
1
So, from above efficiences we can conclude the order as
η
m
I
>
η
I
I
>
η
t
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