A piece of ice of mass `100 g` and at temperature `0^@ C` is put in `200 g` of water of `25^@ C`. How much ice will melt as the temperature of the water reaches `0^@ C` ? (specific heat capacity of water `=4200 J kg^(-1) K^(-1)` and latent heat of fusion of ice `= 3.4 xx 10^(5) J Kg^(-1)`).

Options

A. 128 g
B. 185.4 g
C. 92.8g
D. 61.8g

Answer

D

Explanation

d) Given that, mass of ice-piece, `m_(1)`=100g = 0.1kg
Mass of water, `m_(2)`=200 g=0.2kg
When temperature of mixture is `0^(@)`C, then heat lost by water
`Q=m_(2)c(Deltatheta)=0.2 xx4200(25-0)` (`therefore` specific heat of water = 420 J]
`rArr` Q = 21000 J
Now, mass of melted ice,
`m^(') = Q/L = (21000)/(3.4 xx 10^(-5))`= 61.8 g