8.41

[Aya Shokair- Dis 2H]

A 50.0-g ice cube at 0.0 C is added to a glass containing
400.0 g of water at 45.0 C. What is the fi nal temperature of
the system (see Tables 8.2 and 8.3)? Assume that no heat is lost
to the surroundings.

I understand that q(ice)=-q(water)
So i imagined that we would solve it like this
m(ice) x C(s,ice) x change in temp (ice) = - m(water) x C(s, water) x change in temp (water)

But in the solutions they did something different, and I'm confused.

[Lucian1F]

The temperature of the water is higher than 0 degrees Celsius so you need to take the amount of energy it takes to melt the given amount of ice first and consider that in your calculations.

[Payton Schwesinger 1J]

On the left side of your equation you have to also account for the amount of heat that will be used to melt the ice into water before the temperature equalizes.

[Jessica Jones 2B]

You will need to account for the phase change from ice to water, deltaHfusion.
So your q-ice would include both the mCdeltaT and deltaHfusion.