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VitaX

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## Homework Statement

An 6.0 g ice cube at -12.0˚C is put into a Thermos flask containing 100 cm3 of water at 19.0˚C. By how much has the entropy of the cube-water system changed when a final equilibrium state is reached? The specific heat of ice is 2200 J/kg K and that of liquid water is 4187 J/kg K. The heat of fusion of water is 333 × 103 J/kg.

## Homework Equations

Q = mcΔT

ΔS = Q/T

## The Attempt at a Solution

-Qlost = Qgain -----> -Qw = Qice + Lm

Qw = (.1kg)(4187 J/kg*K)(Tf - 292K) = 418.7Tf - 122,260

Qice = (.006kg)(2200 J/kg*K)(Tf - 261K) = 13.2Tf - 3445

Lm = (333*10^3 J/kg)(.006 kg) = 1998 J

-418.7Tf + 122,260 = 13.2Tf - 3445 + 1998

-431.9Tf = -123,707

Tf = 286.425 K = 13.425 C

Is this the right process so far? If so I'm stuck on what to do next. What comes next in solving this?