## 4C.13

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 4C.13

An ice cube of mass 50.0 g at 0.0 degreesC is added to a glass containing 400.0 g of water at 45.0 degreesC. What is the final temperature of the system (see Tables 4A.2 and 4C.1)? Assume that no heat is lost to the surroundings.

I know that you use the equation q(ice) = -q(water) since the heat lost by the water is absorbed by the ice. So, why can't you just calculate the heat of the ice using q=nC(specific heat capacity of ice)$\Delta$T? Why do you have to calculate in the phase change of the ice to liquid and the increase in temperature instead of just calculating the heat that the ice absorbed from the hot water?

andrewcj 2C
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

### Re: 4C.13

The transfer of heat only stops when both objects are at the same temperature. This means the ice will continue to gain heat until it is melted and at the same temp as the water, or the water has frozen over and is the same temp as the ice. Since I'm assuming the water won't freeze, energy will be continuously transferred to the ice, which will cause it to melt before heating up to match the water's temp.