An 11 g ice cube at -11.0?C is put into a Thermos flask containing 100 cm^3 of w

Question

An 11 g ice cube at -11.0?C is put into a Thermos flask containing 100 cm^3 of water at 17.0?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 × 1063 J/kg.

Explanation / Answer

Q = mc?T ?S = Q/T 3. 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

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