ICSE Class 10 Physics 2018 Solved Paper
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Question : 68
Total: 78
The temperature of 170 g 50 ∘ C 5 ∘ C
Given : Specific heat capacity of water= 4200
J kg − 1 ∘ C − 1 = 336000 J kg − 1
Given : Specific heat capacity of water
Solution:
Given : mass of the water = 170 g =
kg
Initial temperature= 50 ∘ C
Let the mass of ice added be= x kg
Given that the final temperature of mixture= 5 ∘ C
Heat lost by water= m × c × ∆ T
=
× 4200 × ( 50 − 5 )
= 32130 J
Now the change in ice will be as
Ice at0 ∘ C → 0 ∘ C → 5 ∘ C
∴
= m ′ L + m ′ c ∆ T ′
= x × 336000 + x × 4200 × ( 5 − 0 )
= 336000 x + 21000 x
= 357000 x J
By the principle of calorimetry,
Heat lost by water=
∴ 32130 = 357000 x
orx =
= 0.09 kg or 90 g
Initial temperature
Let the mass of ice added be
Given that the final temperature of mixture
Heat lost by water
Now the change in ice will be as
Ice at
By the principle of calorimetry,
Heat lost by water
or
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