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Test Index
AIEEE 2012 Solved Paper
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Section:
Physics
Share question:
© examsnet.com
Question : 37 of 89
Marks:
+1
,
-0
A Carnot engine, whose efficiency is
40
%
40\%
40%
, takes in heat from a source maintained at a temperature of
500
K
500\,\text{K}
500
K
. It is desired to have an engine of efficiency
60
%
60\%
60%
. Then, the intake temperature for the same exhaust (sink) temperature must be:
[AIEEE 2012]
efficiency of Carnot engine cannot be made larger than
50
%
50\%
50%
1200
K
1200\,\text{K}
1200
K
750
K
750\,\text{K}
750
K
600
K
600\,\text{K}
600
K
Validate
Solution:
0.4
=
1
−
T
2
500
0.4=1-\;\frac{T_2}{500}\;\;
0.4
=
1
−
500
T
2
and
0.6
=
1
−
T
2
T
1
\;\;0.6=1-\;\frac{T_2}{T_1}
0.6
=
1
−
T
1
T
2
on solving we get
T
2
=
750
K
T_2=750\,\text{K}
T
2
=
750
K
© examsnet.com
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