Question

A 31 g ice cube at -15.0oC is placed in 139 g of water at 48.0oC. Find the final temperature of the system when equilibrium is reached. Ignore the heat capacity of the container and assume this is in a calorimeter, i.e. the system is thermally insulated from the surroundings. Give your answer in oC with 3 significant figures.
Specific heat of ice: 2.090 J/g K
Specific heat of water: 4.186 J/g K
Latent heat of fusion for water: 333 J/g

Answer 1

Heat energy = mass * specific heat * ∆T
AND
Heat energy = mass * heat of fusion

specific heat of ice = 2.09 J/(g * ˚C)
Heat of fusion if ice = 333 J/g
specific heat of water = 4.186 J/(g * ˚C)

Heat energy will flow from the hot water to the cold ice, until the temperature of the ice = 0. Then heat energy will flow from the water to the ice until all the ice is melted. Then heat energy will flow from the hot water to the cold water, until the temperature of all the water is the same! Since we do not know whether the hot water can melt all the ice, we better determine the temperature of the hot water after the ice is melted!

1st the temperature of the ice increases from -15.0 to 0
Heat energy = 31 * 2.09 * 15.0 = 971.85 J

2nd the ice melts
Heat energy = 31 * 333 = 1.0323 J
Total heat energy = 971.85 + 10323 = 11294.85 J

At this point, all the ice has melted. Now you have 31 grams of water at 0˚. The temperature of the 131 grams of hot water has decreased as the ice warmed up and melted. Let’s determine the temperature of the hot water.
11294.85 = 131 * 4.186 * ∆ T
11294.85 = 548.366 * ∆ T
∆ T = 20.60˚C
The temperature of the hot water = 48 – 20.6 = 27.4˚C

This is the temperature of the 139 grams of hot water

3rd the temperature of the water increases
In the 3rd phase, temperature of the cold water increases as the temperature of hot water decreases, until all the water is at the same temperature.

Now the temperature of the 31 grams of 0˚ water increases and the temperature of the 139 grams of 27.4˚C water decreases, until the temperature of all the water is the same, Tf.

31 * 4.186 * (T_f – 0) = 139 * 4.186 * (27.4 – T_f)
Divide both sides by 4.186

31 * T_f = 139 * (27.4 – T_f)
31 * T_f = 3808.6 – 139 * T_f
170 * T_f = 3808.6
T_f = 22.4˚C

This is the final temperature of the water!