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JEE Main 26-Aug-2021 Shift 1 Solved Paper
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© examsnet.com
Question : 6
Total: 90
An electric appliance supplies
6000
J
∕
min
heat to the system. If the system delivers a power of
90
W
. How long it would take to increase the internal energy by
2.5
×
10
3
J
?
[26 Aug 2021 Shift 1]
2.5
×
10
2
s
4.1
×
10
1
s
24
×
10
3
s
2.5
×
10
1
s
Validate
Solution:
Given, heat supplied to the system,
Δ
Q
Δ
t
=
6000
J
∕
min
=
6000
60
J
∕
s
=
100
J
∕
s
Power delivered,
P
=
Δ
W
t
=
90
W
Increase in internal energy,
Δ
U
=
2.5
×
10
3
J
From first law of thermodynamics, we have
Δ
Q
=
Δ
U
+
Δ
W
or
Δ
Q
Δ
t
=
Δ
U
Δ
t
+
Δ
W
Δ
t
...(i)
Substituting the given values in Eq. (i), we get
100
=
2.5
×
10
3
Δ
t
+
90
⇒
10
=
2.5
×
10
3
Δ
t
⇒
Δ
t
=
2.5
×
10
3
10
⇒
Δ
t
=
2.5
×
10
2
s
© examsnet.com
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