Question

A large punch bowl holds 3.35 kg of lemonade (which is essentially water) at 20.0° C.
A 1.62-kg ice cube at. −10.2°. C is placed in the lemonade.
What is the final temperature of the system, and the amount of ice (if any) remaining?
Ignore any heat exchange with the bowl or the surroundings.

Answer 1

using heat balance, mass times heat capacity times temperature difference, the balance looks like:

3.35 * 1000 g * 4.18 J / g C * (20-T) = 1.62 * 1000 g* 80 * 4.1858 J /g + 1.62 * 1000 g * 4.18/2  J / g C * (10.2 C) + 1.62 * 1000 g * 4.18 J /gC * (T-0)

Final temperature, T is equal to 14.29 C