Specific Heat Problem and Thermos

[RTucekr]

Homework Statement

An insulated Thermos contains 127 cm3 of hot coffee at 79°. You put a 12.0 g ice cube at its melting point to cool the coffee. By how many degrees has your coffee cooled once the ice has melted and equilibrium is reached? Treat the coffee as though it were pure water and neglect energy exchanges with the environment.

mw=0.127kg
Tiw=79+273=352K
mi=0.012kg
Tii=273K

Homework Equations

Q=mcΔT
Q=mLf

The Attempt at a Solution

Qtotal=Qwater+Qice melting+Qice warming=0

mILf+mIcw(Tf-Ti)+mwcw(Tf-Ti)=0
(0.012kg)(333x10^3 J/kg) + (0.012kg)(4180 J/(kg*k))(Tf-273K)+(0.127kg)(4180 J/(kg*K))(Tf-352)=0
Tf=338.303K or 65.302 C

ΔT=79-65.302=13.697C

Any help would be appreciated. I have no idea where I am going wrong with this one.

 

[Shooting Star]

Break it up into two steps. M= mass of water and m = mass of ice.

1. Mass m of ice melts, absorbing mL units of energy, where L is the latent heat of fusion.
Temp of mass M of water drops to certain t1 from 79 C by giving out mL amount of heat. t1 can be found.

2. Mass m of water heats up from 0 C to t2, and mass M of water cools down to t2 from t1. Find t2.

If you use cgs units for this one, take 80 cal/g for L.

 

[ogb p]

Your solution looks correct. The only thing I would suggest is to use the specific heat capacity for water at its liquid state (4180 J/(kg*K)) instead of its solid state (333x10^3 J/kg). This may have been a typo, but it would change your final result slightly to Tf=338.324K or 65.324°C. Other than that, your solution is correct and you have correctly used the equations for heat transfer and phase change. Great job!