(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

Q=mcΔT

Q=mLf

3. 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.

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# Homework Help: Specific Heat Problem and Thermos

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