Question

. Ice at −11.0 °C and steam at 146 °C are brought together at atmospheric pressure in a perfectly insulated container. After thermal equilibrium is reached, the liquid phase at 48.0 °C is present. Ignoring the container and the equilibrium vapor pressure of the liquid, find the ratio of the mass of steam to the mass of ice. The specific heat capacity of steam is 2020 J/(kg.C°).

Answer 1

Using Energy conservation:

Heat gained by ice = Heat released by steam

Q1 + Q2 + Q3 = Q4 + Q5 + Q6

Q1 = Mi*Ci*dT1 = energy required to change -11.0 C ice into water

Q2 = Mi*Lf = energy required to change ice into water

Q3 = Mi*Cw*dT2 = energy required to change 0 C water into 48 C water

Q4 = Ms*Cw*dT3 = energy required to change 100 C water into 48 C water

Q5 = Ms*Lv = energy required to change steam into water

Q6 = Ms*Cs*dT4 = energy required to change 146 C steam into 100 C steam

Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2 = Ms*Cw*dT3 + Ms*Lv + Ms*Cs*dT4

Mi = Mass of ice & Ms = mass of steam

from above expression

Mi*(Ci*dT1 + Lf + Cw*dT2) = Ms*(Cw*dT3 + Lv + Cs*dT4)

Ms/Mi = (Ci*dT1 + Lf + Cw*dT2)/(Cw*dT3 + Lv + Cs*dT4)

Now known values are:

dT1 = 0 - (-11) = 11, dT2 = 48 - 0 = 48 C, dT3 = 100 - 48 = 52 C, dT4 = 146 - 100 = 46 C

Ci = Specific heat of ice = 2090 J-kg/C

Cw = Specific heat of water = 4186 J-kg/C

Cs = Specific heat of steam = 2020 J-kg-C

Lf = latent heat of fusion = 3.34*10^5 J/kg

Lv = latent heat of vaporization = 2.25*10^6 J/kg

Using these values:

Ms/Mi = (2090*11 + 3.34*10^5 + 4186*48)/(4186*52 + 2.25*10^6 + 2020*46)

Ms/Mi = 0.218