Ice heat gain = coffee heat loss.
Assume all the ice melts.
500(g)*80(cal/g)= 500 (g)* 1(cal/gC)*(80-T)
where T is the final temperature.
T = 0
Each gram of melting ice gains absorbs 80 cal of heat while melting. That is enough to lower the temperature of 80 g of water by 1 degree.
In this case, all of the ice melts and the water temperature reaches 0 C.
It the computed T had been less than 0, you would have had to assume that all of the ice did not melt, and do the problem differently.
A physical science student adds 500g of 0 degree C ice to 500g of 80 degree C coffee. What is the final temperature of the drink after thermal equilibrium is reached? How much of the ice was melted by the process?
Assume coffee has same specific heat as water.
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