Question

In a thermally isolated environment, you add ice at 0°C and steam at 100°C. (a) Determine the amount of steam condensed (in g) AND the final temperature (in °C) when the mass of ice and steam added are respectively 84.0 g and 10.8 g.

(b)

Repeat this calculation, when the mass of ice and steam added are interchanged. (Enter the amount of steam condensed in g and the final temperature in °C.)

Answer 1

Part A.

(Simply think that amount of ice is much higher than amount of steam, So there is possibility that all the steam will be condensed, So suppose final temperature is T (less than 100 C). Now Using energy conservation:

Q1 + Q2 = Q3 + Q4

Mi = mass of ice = 84.0 gm = 0.084 kg

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

Q1 = energy absorbed by ice during phase change from ice to water = Mi*Lf = 0.084*3.34*10^5 = 28056 J

Ms = mass of steam = 10.8 gm = 0.0108 kg

Lv = latent heat of vaporization of steam = 2.26*10^6 J/kg

Q3 = energy released by steam during phase change from steam to water = Ms*Lv = 0.0108*2.26*10^6 = 24408 J

(Since Q3 < Q1, So all the amount of steam is condensed)

Amount of steam condensed = 10.8 gm

Q2 = heat absorbed by ice from 0 to T C = Mi*Cw*dT2

Cw = Specific heat capacity of water = 4186 J/kg

dT2 = T - 0 = T

Q2 = 0.084*4186*T

Q4 = heat released by steam from 100 C to T C = Ms*Cw*dT4

dT4 = 100 - T

Q4 = 0.0108*4186*(100 - T)

Using above values:

Q1 + Q2 = Q3 + Q4

28056 + 0.084*4186*T = 24408 + 0.0108*4186*(100 - T)

T = [24408 + 0.0108*4186*100 - 28056]/(0.084*4186 + 0.0108*4186)

T = 2.2 C = final temperature of system

Part B.

(Simply think that amount of ice is much lower than amount of steam, So there is possibility that all the steam will not be condensed, So suppose final temperature is 100 C. Now Using energy conservation:

Q1 + Q2 = Q3

Mi = mass of ice = 10.8 gm = 0.0108 kg

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

Q1 = energy absorbed by ice during phase change from ice to water = Mi*Lf = 0.0108*3.34*10^5 = 3607.2 J

Q2 = heat absorbed by ice from 0 to 100 C = Mi*Cw*dT2 (We've assumed that final temperature is 100 C)

Cw = Specific heat capacity of water = 4186 J/kg

dT2 = 100 - 0 = 100

Q2 = 0.0108*4186*100 = 4520.88 J

Ms = mass of steam = 84.0 gm = 0.084 kg

Lv = latent heat of vaporization of steam = 2.26*10^6 J/kg

Q3 = energy released by steam during phase change from steam to water = Ms*Lv = 0.084*2.26*10^6 = 189840 J

(Since Q3 > Q1 + Q2, So all the amount of steam will not be condensed)

Final temperature of system = 100 C

Now suppose 'm' amount of steam is condensed, then

Q3 = energy released by steam during phase change from steam to water = m*Lv = m*2.26*10^6

Using above values:

Q1 + Q2 = Q3

3607.2 + 4520.88 = m*2.26*10^6

m = (3607.2 + 4520.88)/(2.26*10^6) = 0.0036 kg = 3.6 gm

m = mass of steam condensed = 3.6 gm

T = final temperature of system = 100 C

Let me know if you've any query.