Question

A 81-g ice cube at 0°C is placed in 930 g of water at 30°C. What is the final temperature of the mixture?

Answer 1

Solution-

Heat of fusion of ice = 334 J/g
Specific heat of water = 4.186 J/g*˚C
Amount of heat energy required to melt all the ice.
The water’s temperature will decrease as the ice melts.
Energy absorbed by ice = mass * Heat of fusion = 81 * 334 = 27054 Joules

Now the temperature of the water, after releasing the heat energy.

Energy released by water = Mass * Specific heat * ∆T
Energy released by water =930 * 4.18 * ∆T

930 * 4.186 * ∆T = 81 * 334
∆T = 6.95˚
The water’s temperature decreased 6.95˚ as all the ice melted.

Final temperature of water = 30 = 6.95 = 23.05˚
Now you have 665 grams of water at 23.05˚C and 81 grams of water at 0˚

At last determine the final temperature of the water.
81 * 4.186 * (Tf – 0) = 930 * 4.186 * (23.05 – Tf)
Dividing both sides by 4.186
81 * Tf = 930 * (23.05 – Tf)
1011 * Tf = 21436.5
Tf = 21.20˚ Answer