What Is the Entropy Change When Ice Melts in Hot Water?

[tom4real]

A closed, well-insulated container is filled with 454 g of water at 94.4 °C. To the hot water, 200 g of water ice at exactly 0 °C is added. The mixture reaches an equilibrium temperature of 41.1 °C. Assume the molar heat capacity is constant and all the processes are at constant pressure. The standard enthalpy of fusion for water at 0 °C is 6.008 kJ mol–1. The constant-pressure heat capacity for water is 75.291 J K–1 mol–1. Water has a molecular weight of 18.015 g mol–1.

Calculate the entropy change (in J K–1) for the system that happened because of this mixing.

I know the entropy change equals to q/t because q equals to the enthalpy exchange in the system as it is constant pressure, so what I did was:

q=(454/18.015)75.291(41.1-94.4)+(200/18.015)75.291(41.1)+(200/18.015)*6008=172.2J

change in entropy=172.2/(41.4+273)=0.55 Jk-1

That is apparently incorrect, what have I done wrong?

 

[I like Serena]

A closed, well-insulated container is filled with 454 g of water at 94.4 °C. To the hot water, 200 g of water ice at exactly 0 °C is added. The mixture reaches an equilibrium temperature of 41.1 °C. Assume the molar heat capacity is constant and all the processes are at constant pressure. The standard enthalpy of fusion for water at 0 °C is 6.008 kJ mol–1. The constant-pressure heat capacity for water is 75.291 J K–1 mol–1. Water has a molecular weight of 18.015 g mol–1.

Calculate the entropy change (in J K–1) for the system that happened because of this mixing.

I know the entropy change equals to q/t because q equals to the enthalpy exchange in the system as it is constant pressure, so what I did was:

q=(454/18.015)75.291(41.1-94.4)+(200/18.015)75.291(41.1)+(200/18.015)*6008=172.2J

change in entropy=172.2/(41.4+273)=0.55 Jk-1

That is apparently incorrect, what have I done wrong?

Hi tom4real!

The entropy change is actually:
$$dS=\frac{dQ}{T}$$
When changing temperature at constant pressure and with a constant-pressure heat capacity $C_p$ that means:
$$\Delta S = \int_{T_{initial}}^{T_{final}} \frac{dQ}{T} = \int_{T_{initial}}^{T_{final}} \frac{nC_p\,dT}{T} =nC_p \ln T\Big|_{T_{initial}}^{T_{final}}$$

 

[Chestermiller]

Hi tom4real!

The entropy change is actually:
$$dS=\frac{dQ}{T}$$

Hi I like Serena. In my judgment, this equation needs to be qualified a little by writing $$dS=\frac{dQ_{rev}}{T}$$ where the subscript "rev" refers to a reversible path between the initial thermodynamic equilibrium state and the final thermodynamic equilibrium state. The actual path of this process is not reversible, and, during the irreversible change under consideration, the temperature T of the liquid water is not uniform spatially. So there is no unique temperature T to use in the equation if we don't require the path to be reversible. Only if the path is reversible will the temperature of the water be uniform spatially.

An acceptable reversible path would be to put the 454 grams of water into contact with a continuous sequence of constant temperature reservoirs (running from 94.4 C to 41.1 C), each at a slightly lower temperature than the present temperature of the water. The equation you wrote would then be appropriate for this reversible path. For more discussion of this (with a very similar example), see my recent Physics Forums Insights article at the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/