Problem 4C.13

[Jessa 2K]

I don't understand how to do this problem. Here's what I did:
Ice gains heat; therefore, q is +: qice = (50g)(2.03 J/(C x g))(Tfinal-0)
qice = 101.5 J x Tfinal/C
Water loses heat; therefore q is -: qwater = -(400g)(4.184 J/(C x g))(Tfinal-45)
qwater = -1673.6 J x Tfinal/C + 75,312 J/C
Set qice = qwater
101.5 J x Tfinal/C = -1673.6 J x Tfinal/C + 75,312 J/C
Tfinal = 42.4 Celsius
Thanks!

[Giuliana_Ming_2D]

Hi! You have the qwater correct, but the terms for qice is going to be the amount of energy it takes to melt the ice+the amount of energy it takes to raise the temperature to the final temperature. You get this by splitting the equation into two terms, one is the moles of water multiplied by the standard enthalpy of fusion of water and the other is mcdeltaT, which is (50.0)(4.148)(T-0). Then you set those two terms to qwater and solve for T. Hope this helps!