Question

(a) How much ice at -14.2 C must be placed in 0.150 kg of water at 18.4 C to cool the water to 0.0 C and melt all of the ice?

___0.0318__ kg

(b) If half that amount of ice is placed in the water, what is the final temperature of the water?

____????___ degrees Celcius

Answer 1

Solution of part B.

Suppose the final temperature is T.

Now Using energy conservation:

Heat absorbed by ice = Heat released by water

Q1 = Q2

Q1 = Total heat absorbed by ice = q1 + q2 + q3

Q1 = Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2

dT1 = Tf - Ti = 0 - (-14.2) = 14.2

dT2 = T - 0 = T

Mi = mass of ice = 0.0318/2 = 0.0159 kg

Ci = 2090 J/kg-C & Cw = 4186 J/kg-C

Lf = 3.34*10^5 J/kg

Similarly,

Q2 = heat released by water from 18.4 C to T C = Mw*Cw*dT3

dT3 = 18.4 - T

Mw = mass of water = 0.150 kg

Now using given values:

Q1 = Q2

0.0159*2090*14.2 + 0.0159*3.34*10^5 + 0.0159*4186*T = 0.150*4186*(18.4 - T)

Now Solving above equation

T = [0.150*4186*18.4 - (0.0159*2090*14.2 + 0.0159*3.34*10^5)]/(0.0159*4186 + 0.150*4186)

T = 8.3 C

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