Phy heat question

Specific latent heat of fusion of ice, L = 334000 J kg-1
Specific heat capacity of water, c = 4200 J kg oC-1

Let m kg be the mass of ice in 0.1 kg of the mixture of ice and water.
Let 0oC be the initial temperature of the mixture of ice and water.

Heat absorbed for (0oC ice → 0oC water → 10oC water)
= mL + mc∆T
= m(334000) + 0.1(4200)(10 - 0)
= (334000m + 4200) J

Heat released for (40oC water → 10oC water)
= mc∆T
= (0.15)(4200)(40 - 10)
= 18900 J

Assume that there is no heat loss to the surroundings.
Heat absorbed = Heat released
334000m + 4200 = 18900
334000m = 14700
m = 0.044 kg

Mass of ice in the mixture = 0.044 kg
=