ICSE Class 10 Physics 2017 Solved Paper
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Question : 34
Total: 65
Calculate the mass of ice needed to cool 150 g 50 g 32 ∘ C 5 ∘ C
Specific heat capacity of calorimeter= 0.4 J ∕ g C C
Specific heat capacity of water= 4.2 J ∕ g ∘ C
Latent heat capacity of ice= 330 J ∕ g
Specific heat capacity of calorimeter
Specific heat capacity of water
Latent heat capacity of ice
Solution:
Let the mass of ice be = ′ m ′ g
As ice will gain heat from calorimeter and water, it will undergo conversion as :
Ice → Water → Water
(at 0 ∘ C ) (at 0 ∘ C ) (at 5 ∘ C )
So, heat gained= m L + m c ∆ T
= m × 330 + m × 4.2 × ( 5 − 0 )
= 330 m + 21 m
= ( 351 m ) J .
∵
∴
= m 1 c 1 ∆ T 1
= 50 × 0.4 × ( 32 − 5 )
= 540 J
Heat lost by hot water
= m 2 C W ∆ T 2
= 150 × 4.2 × ( 32 − 5 )
= 150 × 4.2 × 27
= 17010 J
∴ = 540 + 17010 = 17550 J
By the principle of calorimetry
Heat gained = Heat lost
∴ 351 m = 1 7 5 5 0
or m =
= 50 g
As ice will gain heat from calorimeter and water, it will undergo conversion as :
So, heat gained
Heat lost by hot water
By the principle of calorimetry
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