Question

John imagines himself on the beach playing some volleyball once Memorial Day comes around. After a victorious game, John always dreams of a cold cup of water but as his future holds he only has 40.0 g of water at room temperature (25°C).

a. How much heat will be released if all the water, 40 g at 25°C, is converted to ice at 0°C.

b. If John still has his original cup of 40.0 g of water at room temperature (25°C), determine the composition and temperature of the mixture if John adds 30.0 g of ice at -10.0°C into the liquid water? Show how you arrived to this conclusion.

Answer 1

Part A.

If all available water is converted into ice, then heat released will be:

Q = Q1 + Q2

Q1 = Heat released by water from 25 C to 0 C = Mw*Cw*dT

Q2 = Heat released by phase change from water to ice = Mw*Lf

Q = Mw*Cw*dT + Mw*Lf

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

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

Mw = mass of water = 40 gm = 0.040 kg

So,

Q = 0.040*4186*25 + 0.040*3.34*10^5

Q = 17546 J = Amount of heat released (Use -ve sign for heat released if sign requirement is mentioned in your system)

Part B.

Suppose final temperature of water + ice mixture is T C, then (Here we've assumed that heat is not transferred into environment or into the cup)

Heat absorbed by ice = Heat released by water

Q1 + Q2 = Q3

Q1 = Heat absorbed by ice from -10 C to 0 C = Mi*Ci*dT1

Q1 = 0.030*2090*10 = 627 J

Q2 = Heat absorbed during phase change = Mi*Lf = 0.030*3.34*10^5 = 10020 J

Q3 = Heat released by water from 25 C to 0 C = Mw*Cw*dT3 = 0.040*4186*25 = 4186 J

Since Q1 + Q2 > Q4, So all the ice will not be change into water and final equilibrium temperature will be 0 C

Now suppose m amount of ice is converted into water, then

Amount of heat available for phase change step = Q3 - Q1 = 4186 - 627 = 3559 J

Q2 = m*Lf

m = Q2/Lf = 3559/(3.34*10^5) = 0.0107 = 10.7*10^-3 kg = 10.7 gm

So 10.7 gm of ice will be converted into water from ice

final composition will be:

mass of remaining ice = 30 gm - 10.7 gm = 19.3 gm

mass of water in cup = 40 gm + 10.7 gm = 50.7 gm

Let me know if you've any query.