Question

(a) Two 64 g ice cubes are dropped into 278 g of water in a thermally insulated container. If the water is initially at 25°C, and the ice comes directly from a freezer at −15°C, what is the final temperature at thermal equilibrium? (in celcius)

(b) What is the final temperature if only one ice cube is used? (in celcius)

Answer #1

Heat released by ice =

Q1 = Q3 + Q4

Q1 = m1*Cp*dT + m1*Lf

Q1 = 0.128*2030*15 + 0.128*3.36*10^5

Q1 = 46905.6 J

Heat absorbed by water:

Q2 = m2*Cp*dT2

Q2 = 0.278*4186*25 = 29092.7

Q1 > Q2 & Q2 > Q3

It means there is not enough heat to melt complete ice.

So Temp. at thermal equilibrium = 0 C

B.

Q1 = Q3 + Q4

Q1 = 0.064*2030*15 + 0.064*3.34*10^5 = 23324.8 J

Q2 = 0.278*4186*25 = 29092.7

Q2 > Q1, So all the ice will melt

Now temperature at thermal equilibrium will be

T = (0.278*4186*25 - 0.064*2030*15 - 0.064*3.34*10^5)/(0.278*4186 + 0.064*4186)

T = 4.03 C

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