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Byju's Answer
Standard XII
Chemistry
Second Law of Thermodynamics
A heat engine...
Question
A heat engine is operating between
500
K
to
300
K
and it absorbs
10
k
c
a
l
of heat from
500
K
reservoir reversibly per cycle. Calculate the work done (in kcal) per cycle.
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Solution
Heat rejected at the higher temperature is -
Q
2
=
T
2
T
1
×
Q
1
=
300
500
×
10
=
6
k
c
a
l
So, work done
=
Q
1
−
Q
2
=
10
−
6
=
4
k
c
a
l
.
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