When 0.15 kg of ice of `"0 "^(@)C` mixed with 0.30 kg of water at `"50 "^(@)C` in a container, the resulting temperature is `6.7" "^(@)C`. Calculate the heat of fusion of ice. `(s_("water")=4186" J Kg"^(-1)K^(-1))`

Explanation

Mass of ice `m_(1)=0.15` kg
Temperature of ice `t_(1)=0^(@)C`
Mass of water `m_(2)=0.30` kg
Temperature of water `t_(2)=50^(@)C`
Temperature of mixture of water and ice `t_(3)=6.7^(@)C`
Specific heat of water `S=4186" Jkg"^(-1)K^(-1)`
Expose latent heat of fusion of ice is `L_(f)`
`rArr` Heat required to melt the ice,
`Q_(1)=m_(1)L_(f)`
`:.Q_(1)=0.15L_(f)J" . . . (1)"`
`rArr` Heat loss by water,
`Q_(2)=m_(2)S(t_(2)-t_(3))`
`=0.30xx4186(50-6.7)`
`=0.30xx4186xx43.3`
`:.Q_(2)=54376.14J` . . .(2)
Heat required to make temperature of water of ice `6.7^(@)C`,
`Q_(3)=m_(1)S(t_(3)-t_(1))`
`=0.15xx4186(6.7-0)`
`=0.15xx4186xx6.7`
`:.Q_(3)=4206.93J` . . . (3)
{Heat lost by water} = {Heat required to melt the ice} + {Heat required to make temperature of water to ice `t_(3)`}
`Q_(2)=Q_(1)+Q_(3)`
`:.54376.14=0.15L_(f)=4206.96" "` [from equation (1), (2) and (3)]
`:.0.15L_(f)=54376.144206.96`
`=50xx169.21`
`:.L_(f)=(50xx169.21)/(0.15)=334461.4" Jkg"^(-1)~~3.34xx10^(5)" Jkg"^(-1)`