Questions

80 grams of water at 30° C are poured on a large block of ice at 0° C. The mass of ice melted is:(Latent Heat of fusion of ice =80 cal ${g}^{-1}$)

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By Expert Faculty of Sri Chaitanya
a
30 grams
b
80 grams
c
160 grams
d
60 grams
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detailed solution

Correct option is A

from principle of method of mixturesheat lost by water = heat gained by icems ∆T=mice meltedL---(1)here s is specific heat of water =1m is mass of water= 80g (given)∆T is change in temperature= 30-0=300CL is latent heat of fusion of ice =80 calorie g-1  substitute above values in equation (1)80(1)(30)=m(80)mass of ice melted = 30 g

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where $\mathrm{\theta }$ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are  and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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