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.
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?