ICSE Class 10 Physics 2016 Solved Paper
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Question : 39
Total: 73
A copper vessel of mass 100 g 150 g 50 ∘ C 5 ∘ C
Given :
Specific heat capacity of copper= 0.4 Jg − 1 ∘ C − 1
Specific heat capacity of water= 4.2 J g − 1 C − 1
Specific latent heat of fusion of ice= 336 Jg − 1
Given :
Specific heat capacity of copper
Specific heat capacity of water
Specific latent heat of fusion of ice
Solution:
Heat given by water to reach 5 ∘ C + 5 ∘ C = 0 ∘ C 5 ∘ C
m c ∆ t + m c ∆ t = m L + m c ∆ t
where∆ t =
[ 150 × 4.2 × ( 50 ∘ − 5 ∘ ) ] + [ 100 × 0.4 × ( 50 ∘ − 5 ∘ ) ]
= ( m × 336 ) + [ m × 4.2 × ( 5 − 0 ) ] ( 150 × 4.2 × 45 ) + ( 100 × 0.4 × 45 )
= ( m × 336 ) + ( m × 4.2 × 5 )
28350 + 1800 = 336 m + 21 m
357 m = 30150
m =
= 84.45 gm of ice.
where
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