Question

In a well-insulated calorimeter, 1.0 kg of water at 20 ∘ C is mixed with 1.0 g of ice at 0 ∘ C . What is the net change in entropy Δ S sys of the system from the time of mixing until the moment the ice completely melts? The heat of fusion of ice is L f =3.34× 10 5 J/kg . Note that since the amount of ice is relatively small, the temperature of the water remains nearly constant throughout the process. Note also that the ice starts out at the melting point, and you are asked about the change in entropy by the time it just melts. In other words, you can assume that the temperature of the "ice water" remains constant as well. Express your answer numerically in joules per kelvin. Use two significant figures in your answer.

Answer 1

Heat transferred to completely melt the ice

Q = mL

where L is heat of fusion for ice

Q = 1e-3 * 3.34e5

Q = 334 J

so,

Entropy change of ice

S = Q / T

S = 334 / 273.15

S = 1.22277 J/K

Now, for water

Q = mc (Tf - Ti)

Tf = 19.9201 ^oC

Entropy change of water

S_w = 1000 * 4.18 ln ( 293.07 / 293.15)

S_w = - 1.14086 J/K

so,

S (system) = S - S_w

S (system) = 0.082 J/K