ICSE Class 10 Physics 2020 Solved Paper
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Question : 59
Total: 67
A piece of ice of mass 60 g g 50 ∘ C
(Assume no heat is lost to the surrounding)
Specific heat capacity of water= 4.2 Jg − 1 k − 1
Specific latent heat of fusion of ice= 336 Jg − 1
(Assume no heat is lost to the surrounding)
Specific heat capacity of water
Specific latent heat of fusion of ice
Solution:
Given : mass of water ( m ) = 140 g = 50 ∘ C
mass of ice( m ′ ) = 60 g = 0 ∘ C
Let the final temperature= T ∘ C
By the principle of method of mixture, Heat lost by water=
m c ∆ t = m ′ L + m ′ C ∆ t ′
140 × 4.2 × ( 50 − T ) = 60 × 336 + 60 × 4 ⋅ 2 × ( T − 0 )
588 ( 50 − T ) = 20160 + 252 T
29400 − 588 T = 20160 + 252 T
29400 − 20160 = 588 T + 252 T
9240 = 840 T
or T =
= 11 ∘ C .
mass of ice
Let the final temperature
By the principle of method of mixture, Heat lost by water
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