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Chapter 18: Problem 77

A \(74.6-\mathrm{g}\) ice cube floats in the Arctic Sea. The temperature and pressure of the system and surroundings are at 1 atm and \(0^{\circ} \mathrm{C}\). Calculate \(\Delta S_{\text {sys }}, \Delta S_{\text {surr, }}\) and \(\Delta S_{\text {uniy }}\) for the melting of the ice cube. What can you conclude about the nature of the process from the value of \(\Delta S_{\text {univ }} ?\) (The molar heat of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol} .)\)

Short Answer

Expert verified
The change in entropy of the system (\(\Delta S_{sys}\)) is approximately 0.090 kJ/K. The change in entropy of the surroundings (\(\Delta S_{surr}\)) is approximately -0.090 kJ/K. The change in entropy of the universe (\(\Delta S_{univ}\)) is 0 kJ/K. The process of the melting of the ice cube is at equilibrium, thus reversible.

Step by step solution

01

Calculate the Number of Moles of Water

First convert the mass of the ice cube into moles using the known molar mass of water (18.015 g/mol): \[n=\frac{m}{M}=\frac{74.6 \ g}{18.015 \ g \ / \text{mol}} \approx 4.14 \text{mol}\]
02

Calculate ΔSsys

Then calculate the change in the entropy of the system using the equation: \(\Delta S_{sys}=\frac{q_{rev}}{T}=\frac{n \Delta H_{fus}}{T}\). Substitute the values (note that the molar heat of fusion ΔHfus and temperature T must be in the same units):\[\Delta S_{sys} = \frac{4.14 \text{mol} \times 6.01 \text{kJ/mol}}{273.15K} \approx 0.090 \text{kJ/K}\]
03

Calculate ΔSsurr

The ΔSsurr is equal to the negative of ΔSsys as the melting process is reversible at equilibrium: \[\Delta S_{surr} = - \Delta S_{sys} = - 0.090 \text{kJ/K}\]
04

Calculate ΔSuniv

Then sum up \(\Delta S_{sys} + \Delta S_{surr}\) to find \(\Delta S_{univ}\):\[\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} = 0 \text{kJ/K}\]
05

Interpret the Results

Since \(\Delta S_{univ}\) equals 0, the process of melting is concluded to be at equilibrium. Thus, it is a reversible process.

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