Question

What is the final equilbrium temperature when 40.0 grams of ice at -12.0 degrees C is mixed with 20.0 grams of water at 32 degrees C? The specific heat of ice is 2.10 kJ/kg degrees C, the heat of fusion for ice at 0 degrees C is 333.7 kJ/kg, the specific heat of water 4.186 kJ.kg degrees C, and the heat of vaporization of water at 100 degrees C is 2,256 kJ/kg.

A. How much energy will it take to cool water to 0 degrees C? How much energy will it take to warm the ice to 0 degrees C?

B. Which will reach 0 degrees C, the ice or the water? What temperature will the other material reach?

C. If all the ice melts. what will the final temperature of all the water be? If it doesnt all melt, how much ice and how much water will there be? What temperature will they be at?

Answer 1

given

mi = 40 g

Ti = -12 C

mw = 20 g

Tw = 32 C

Ci = 2100 J/kg C

Lf = 333700 J/ kg

Cw = 4186 J/kg C

a. energy required to cool water to 0 C = Ec

Ec = mw*Cw*(0 - Tw) = -2679.04 J

energy required to warm the ice to 0 C = Ei

Ei = mi*Ci(0 - Ti) = 1008 J

b. hence as ice requires less energy, it will cool to 0 C first, the water reaches temperature T

1008 = mw*Cw(32 - T)

T = 19.959 C

c. if all ice melts, Energy released = E

E = mi*Lf = 13348 J

this energy is a lot more than the energy required to bring water to 0 C

hence all of ice will not melt

final temperaure of water = 0 C