Final Temperature?

Question

Determine the final temperature Tf, that results when 43 grams of ice at 0 °C are mixed with 300 grams of liquid water at 50 °C.
Specific heat of water: c = 1.00 cal/(gram*°C).
Heat of fusion for the ice - liquid water transition: cF = 79.7 cal/gram.

Answer 1

First, the water will melt the ice. So calculate how much energy that requires. Recall, Q = m*(cF), or

Q = 43 * 79.7 = 3427.1 cal

this heat needs to come from the water and will lower the water temperature. Here, Q = mc*(delta T), or

3427.1 = 300 *1 * (50 - T)

Solve for T = 38.57 degree C

Now there is 300 grams of water at 38.57 degree and 43 grams at 0 degrees. These will move to the same temperature by exchanging heat. Again Q = mc*(delta T) with Q equal to the amount of heat given up by the warm water and passed to the cold water. This means

43 *1 *(Tf - 0) = 300 *1 * (38.57 - Tf)

Solve for Tf = 33.74 degrees C