problem about resulting temperature?

Question

if 50g of ice were put into 200g of water at 75degrees celsius, what is the resulting temperature?

Answer 1

Well, that would be something. First we have to determine whether it is possible to melt all of the ice.

Q = mLf
Q = (50 g)(334 J/g)
Q = 16700 J

If water was reduced to 0 degrees Celsius, then

Q = mc(Tf - Ti)
Q = (200 g)(4.19 J/g C)(0 - 100)
Q = -83800 J

Yeah, it is possible, since the "hotter" water is able to give up more heat than the ice can absorb. Thus, we solve it now

Qgained = -Qlost
mL(ice) + mc(Tf - Ti) (ice turned into water) = -mc(Tf - Ti) (water at 75 C)
(50)(334) + (50)(4.19)(Tf - 0) = -(200)(4.19)(Tf - 75)
16700 + 209.5Tf = -838Tf + 62850
1047.5Tf = 46150
Tf = 44.06 C

Note that the mLf part is important as heat causes 2 changes, which cannot occur simultaneously: phase change and temperature change. The ice absorbs heat not only to melt, but also to have its temperature changed.

Well, that's what I got. Hope that helps.