Question

If water is pure enough and in a sufficiently smooth container, it can sometimes be “supercooled” so that it remains liquid for a time even when its temperature is lowered below its freezing point. (For this problem, you can use table 8.2 for heat capacities and may assume that they do not change with temperature.)

(a) A beaker with 500 g liquid water initially at 10 ºC is placed in a freezer at –15 ºC, where the liquid supercools until it is at the same temperature as the freezer. Calculate qH2O, and pay careful attention to its sign.

(b) The heat of fusion of ice at 0 ºC is 6.01 kJ/mol (Table 8.3), but this number changes slightly with temperature. Calculate the heat of fusion of ice at –15 ºC.

(c) The beaker of water from part (a) (now at –15 ºC) is taken out of the freezer and jostled, which triggers the water to freeze rapidly (so quickly that the beaker is effectively insulated; no heat is transferred to or from the beaker’s surroundings). As the ice forms, heat is given off and so the temperature of the water/ice mixture rises until it reaches 0 ºC, at which point no more ice forms. How much ice will form during this process?

Answer 1

a) Q = m*Cp*(T2-T1)
calculate Q(H2O), you have to use all the variables of the H2O given. These variables are:
mass of H2O, m = 500 g
specific heat of H2O, Cp = 4.184 J/gC : (not given but can be assumed see in table)
Initial Temp of H2O, T1 = 10 ' C
Final Temp of H2O, T2 = -15 ' C
Putting all this together will yield:
Q = (500 g) * (4.184 J/gC) * (-15 C-10 C) = 52300J

b) heat of fusion (Hf)= hf(Tf - 0°C)

hf = heat of fusion of ice at 0 ºC is 6.01 kJ/mol

Tf =final temperature (-15 C)

Hf =-90.15 j/mol :: { - tive is due to final negetive temperature -15 c so conseder

mangnitude only not sign }

c) m ice×Hf + m ice×Cp×(Tf - 0°C) = m water×Cp×(Tin - Tf)
Where
m ice = weight of ice = ?
Hf = Heat of fusion = 79.76 calories/gram {convert 6.06kj/mol into calories/gram )
m water = weight of water = 500g
Tf = final temperature, =0 c
Ti = initial water temperature = -15°C
Cp = Specific Heat for water = {1.0 calories per gram per °C from table}

Substituting the values we get:
m ice×79.76+ m ice×1×(-15 C - 0°C) = 500×1×(-15°C - 0 C) { -tive temp will take magnitude not sing}
m ice = 7500/94.76
m ice =79.14 g
maximum form of ice is 79.14 gram