Question

12 grams of steam at T = 120° C is added to 55 grams of ice at T = -45° C.

a) Find the final temperature of the system, assuming all ice melts and all steam condenses.

b) If the mass of steam is reduced to 4.2 grams, find the final temperature of the system. If the final temperature is 0° C, determine how much ice melts.

Answer 1

a)

Tfinal

Qsteam= 12*2.03*(100-120) = -487.2

Qcond = 12*-2264 = -27168

Qtotal = (487.2+27168) = -27655.2

now, we get water at 100°C

for ice:

Qice = m*C*(Tf-Ti) = 55*2.01*(0--45) = 4974.75 J

Qmelt = m*LH = 55*334 = 18370 J

Qtotal ice = 18370+4974.75 = 23344.75 J

then...

-Qlost = Qgain

-(-27655.2 + Msteam*Cpwater*(Tf-100)) = 23344.75 + Mice*Cpwater *(Tf-0)

solve

27655.2 - 12*4.184*(Tf-100) = 23344.75 +55*4.184*(Tf)

-50.2Tf + 5020 = 23344.75 -27655.2 +230.12Tf

5020 - 23344.75 +27655.2   = Tf*(230.12+50.2)

Tf = 9330.45/280.32 = 33.284 °C