Question

An insulated Thermos contains 115 g of water at 78.7 ˚C. You put in a 10.0 g ice cube at 0.00 ˚C to form a system of ice + original water. The specific heat of liquid water is 4190 J/kg•K; and the heat of fusion of water is 333 kJ/kg. What is the net entropy change of the system from then until the system reaches the final (equilibrium) temperature?

Answer 1

We know that

To melt the ice requires 2294 J; this will reduce the temperature of the water by
[2294/(4.19*115)] ⁰C = 4.76 ⁰C .

Now we have 10.0 g of water at 0.00 ⁰C and
115 g of water at 78.7 ⁰C  
the equilibrium temperature will be found from
(115)(78.7 - T) = (10.0)_T, yielding

9050.5 - 115T = 10

T = 78.61 ⁰C

Now Lets see Entropy changes
(a) The melting of the ice gives dS = 2294 J/(273.15 K) = +8.40 J/K
(b) The warming of the melted ice gives
dS = (10.0 g)(4.19 J/(g K)) ln[(273.15+78.61)/273.15]
= (41.9 J/K) ln(1.2978) = +10.59 J/K
(c) The cooling of the originally warm water gives
dS = -(115 g)(4.19 J/g K)) ln[(273.15+ 78.7)/(273.15+78.61)]
= -481.85 J/K* ln(1.000255) = -0.122 J/K

So the net entropy change is +8.40 J/K +10.59 J/K -0.122 J/K = 18.868 J/K