Solution:

Given:

Mass of ice, m = 40 g
Initial temperature of ice, Ti = -10°C
Final temperature of mixture, Tf = 80°C

Let's assume that all the steam condenses and transfers its heat to the ice to melt it completely.

Phase Changes:

Steam → Liquid water → Water at 0°C → Water at 80°C
Ice at -10°C → Water at 0°C

Calculation:

The heat required to melt the ice at 0°C:

Q1 = mL
Where,
m is the mass of ice = 40 g
L is the latent heat of fusion for ice = 335 J/g (at 0°C)

Q1 = 40 g x 335 J/g = 13400 J

The heat required to heat the melted ice from 0°C to 80°C:

Q2 = mcΔT
Where,
m is the mass of melted ice = 40 g
c is the specific heat capacity of water = 4.18 J/g°C
ΔT is the change in temperature = (80°C - 0°C) = 80°C

Q2 = 40 g x 4.18 J/g°C x 80°C = 13424 J

The heat required to condense the steam:

Q3 = mL
Where,
m is the mass of steam = mass of water produced from the steam = (mass of ice/1000) = 0.04 kg
L is the latent heat of vaporization for water = 2260 J/g (at 100°C)

Q3 = 0.04 kg x 2260 J/g = 90.4 J

The total heat required:

Q = Q1 + Q2 + Q3

Q = 13400 J + 13424 J + 90.4 J = 26814.4 J

The mass of the mixture:

The mass of the mixture is the sum of the mass of melted ice and the mass of water produced from the steam.

Mass of water produced from the steam = mass of steam = 0.04 kg
Mass of melted ice = 40 g = 0.04 kg

Mass of the mixture = 0.04 kg + 0.04 kg = 0.08 kg

Therefore, the mass of the mixture will be approximately 0.08 kg.

Conclusion:

The mass of the mixture will be approximately 0.08 kg.

M1LF+M1SDT=M2LV
40×80+40×1×(80-0)=m2×540
6400/540=m2
m2=12
m of mixture =40+12=52