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JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 3 Questions
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© examsnet.com
Question : 93
Total: 100
A steam engine intakes
50
g
of steam at
100
∘
C
per minute and cools it down to
20
∘
C
. If latent heat of vaporization of steam is
540
cal
g
−
1
, then the heat rejected by the steam engine per minute is _____
×
10
3
cal
(Given : specific heat capacity of water :
1
cal
g
−
1
∘
C
−
1
)
[25-Jun-2022-Shift-1]
Your Answer:
Validate
Solution:
∆
Q
rej
=
50
×
540
+
50
×
1
×
(
100
−
20
)
=
50
×
[
540
+
80
]
=
50
×
620
=
31000
cal
=
31
×
10
3
cal
© examsnet.com
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