    Question

# A piece of ice of mass 40 g at 0oC is added to 200 g of water at 50∘C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200Jkg−1K−1 and specific latent heat of fusion of ice = 336×103Jkg−1 :

A
28.33K
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B
28.33C
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C
2.833C
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D
28.67C
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Solution

## The correct option is B 28.33∘CLet the final temperature be Tice(0oC) is 40g(40×10−3Kg)water(50oC) is 200g(200×10−3Kg)CASE−IHeat gained by ice at 0oC from converting to water at 0oC=mL=40×10−3×336×103JAlso heat gain by water at 0oC to achieve final temperature T=ms△T=40×10−3×4200×(T−0)∴ Total heat gained by ice at 0oC to change water at ToC=mL+ms△T=(4×42T)+(40×336)=13440+168T(1)CASE−IIHeat lost by water as its temperature decreases from 50oC to ToC=0.2×(4200)(50−T)=4200−840T(2)Heat gain by ice = Heat lost by water ((1)=(2))⇒13440+168T=4200−840T⇒T=28.3oC  Suggest Corrections  0      Similar questions  Related Videos   Calorimetry
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