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An ice cube (mass 30 g) at 0°C is left sitting on the
Chapter 3, Problem 10P(choose chapter or problem)
An ice cube (mass 30 g) at \(0^{\circ} \mathrm{C}\) is left sitting on the kitchen table, where it gradually melts. The temperature in the kitchen is \(25^{\circ} \mathrm{C}\).
(a) Calculate the change in the entropy of the ice cube as it melts into water at \(0^{\circ} \mathrm{C}\). (Don’t worry about the fact that the volume changes somewhat.)
(b) Calculate the change in the entropy of the water (from the melted ice) as its temperature rises from \(0^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\).
(c) Calculate the change in the entropy of the kitchen as it gives up heat to the melting ice/water.
(d) Calculate the net change in the entropy of the universe during this process. Is the net change positive, negative, or zero? Is this what you would expect?
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QUESTION:
An ice cube (mass 30 g) at \(0^{\circ} \mathrm{C}\) is left sitting on the kitchen table, where it gradually melts. The temperature in the kitchen is \(25^{\circ} \mathrm{C}\).
(a) Calculate the change in the entropy of the ice cube as it melts into water at \(0^{\circ} \mathrm{C}\). (Don’t worry about the fact that the volume changes somewhat.)
(b) Calculate the change in the entropy of the water (from the melted ice) as its temperature rises from \(0^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\).
(c) Calculate the change in the entropy of the kitchen as it gives up heat to the melting ice/water.
(d) Calculate the net change in the entropy of the universe during this process. Is the net change positive, negative, or zero? Is this what you would expect?
ANSWER:Step 1 of 4
To calculate the change in the entropy of the ice cube \(\Delta S_{1}\) as it melts into water at \(0^{\circ} \mathrm{C}\)
Mass of the ice cube m = 30 g or 0.03 kg
Temperature of ice cube \(\mathrm{T}_{\mathrm{C}}=0^{\circ} \mathrm{C} \text { or } 273 \mathrm{~K}\)
Temperature of the kitchen \(\mathrm{T}_{\mathrm{H}}=25^{\circ} \mathrm{C} \text { or } 298 \mathrm{~K}\)
The entropy of the ice cube (constant temperature)
\(\Delta S_{1}=\frac{Q_{1}}{T_{C}}\)
Where Q is the energy needed to melt the ice
\(\mathrm{Q}_{1}=\mathrm{m} \times \mathrm{L}_{\mathrm{f}}\)
\(\mathrm{L}_{\mathrm{f}}\) is the latent heat of fusion of ice
\(\mathrm{L}_{\mathrm{f}}=334 \mathrm{KJ} / \mathrm{K}\)
\(\begin{array}{l}
\Delta S_{1}=\frac{m \cdot L f}{T_{C}} \\
\Delta S_{1}=\frac{0.03^{*} 334^{*} 10^{3}}{273} \\
\Delta S_{1}=36.7 \mathrm{~J} / \mathrm{K}
\end{array}\)
The change in the entropy of the ice cube is 36.7 J/K
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