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ICSE Class 10 Answered

A piece of ice of mass 60 g is dropped into 140 g of water at 50°C. Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding) Specific heat capacity of water = 4.2 Jg 'k! Specific latent heat of fusion of ice = 336 Jg!
Asked by ruby19lko | 13 Mar, 2020, 01:42: PM
answered-by-expert Expert Answer
Let final temperature of water is T .

Thermal energy gain by ice = mice × ( L  + Cp T) = 60 × ( 336  + 4.2 × T )
 
where mice is mass of ice,  L is latent heat of fusion of ice and Cp is specific heat of water.

Thermal energy lost by water  = mw × Cp × ( 50 - T ) = 140 × 4.2 × ( 50 - T )
 
where mw is mass of water
 
By conservation of energy, we have, 140 × 4.2 × ( 50 - T ) = 60 × ( 336  + 4.2 × T )
 
from above equation, we get T = 11 oC
Answered by Thiyagarajan K | 13 Mar, 2020, 02:41: PM

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