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Byju's Answer
Standard XII
Physics
1st Law of Thermodynamics
80 g of water...
Question
80
g
of water at
30
o
C
is poured on large block of ice at
0
∘
C. The mass of ice that melts is
Given, Latent heat =
334
J
/
g
Specific heat of water
=
4.2
J
g
−
1
C
o
−
1
A
1600
g
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B
30
g
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C
150
g
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D
80
g
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Solution
The correct option is
B
30
g
Δ
T
=
30
o
C
Heat released by water is abosrbed by ice ,the final temperature of water will be
0
o
C
⇒
8
×
1
×
30
=
m
×
80
⇒
m
=
30
g
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