Calculating Final Temp of Water in Styrofoam Cup

[ramenluver50]

Homework Statement

A 54.0 g ice cube, initially at 0°C, is dropped into a Styrofoam cup containing 327 g of water, initially at 19.4°C. What is the final temperature of the water, if no heat is transferred to the Styrofoam or the surroundings?

Homework Equations

Q=mc(delta)T
mc(delta)T+mc(delta)T = 0
Q=mL
Given- L=3.35e5
C(water) = 4190

The Attempt at a Solution

First i tried,
mc(delta)T+mc(delta)T = 0

.054kg*4190(T-0)+.327kg*4190(T-19.4)=0


then i tried...
mL+mc(delta)T+mc(delta)T = 0

(.054kg)(3.35e5)+.054kg*4190(T-0)+.327kg*4190(T-19.4)=0

both solve for Temp of course.
and both are wrong... help?!

 

[Redbelly98]

The Attempt at a Solution

First i tried,
mc(delta)T+mc(delta)T = 0

.054kg*4190(T-0)+.327kg*4190(T-19.4)=0


then i tried...
mL+mc(delta)T+mc(delta)T = 0

(.054kg)(3.35e5)+.054kg*4190(T-0)+.327kg*4190(T-19.4)=0

both solve for Temp of course.
and both are wrong... help?!

Your second method is correct, since heat is required to melt the ice.

If you share your answer with the rest of us, we could tell you if you are at least close to the correct answer, and it's maybe a roundoff or significant figures problem. Or that you are way off and made an arithmetic mistake.