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Chapter 19: Problem 43

A shallow pond contains 94 Mg of water. In winter, it's entirely frozen. By how much does the entropy of the pond increase when the ice, already at \(0^{\circ} \mathrm{C},\) melts and then heats to its summer temperature of \(15^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The total increase in entropy of the pond when the ice melts and heats to 15°C is given by the sum of \(\Delta S1\) and \(\Delta S2\) as calculated in the steps.

Step by step solution

01

Calculate the heat required to melt the ice

The heat required to melt ice into water at 0°C (Q1), is given by \( Q = m \cdot L_f \), where m is the mass and \( L_f \) is the latent heat of fusion for water, which is approximately \(334 \, kJ/kg\). Convert mass m from Mg to kg. Now, find Q1 = \( 94 \times 10^6 \, kg \times 334 \, kJ/kg \).
02

Calculate the increase in entropy during melting

The increase in entropy (\( \Delta S1 \)) during the melting is given by \( \Delta S = \frac{Q}{T} \). Here, Q is the heat calculated in step 1 and T is the temperature of the phase change which should be in Kelvin. Note that the melting temperature of water (0°C) in Kelvin is 273.15K. Hence, compute \( \Delta S1 = \frac{Q1}{273.15} \). The value will be in \( kJ/K \).
03

Calculate the heat required to heat the water

The heat required to heat the water from 0°C to 15°C (Q2), is given by \( Q = m \cdot c \cdot \Delta T \), where m is the mass, c is the specific heat capacity of water (approximately \(4.186 \, kJ/kg \cdot K\)), and \( \Delta T \) is the change in temperature. Now, find Q2 = \( 94 \times 10^6 \, kg \times 4.186 \, kJ/kg \cdot K \times 15 \, K \).
04

Calculate the increase in entropy during heating

For the second phase of heating the water, the increase in entropy (\( \Delta S2 \)) is calculated by integrating the function \( \frac{dQ}{T} \). However, since the temperature change is small, an average temperature of 7.5°C (280.65K in Kelvin) can be used. Hence, compute \( \Delta S2 = \frac{Q2}{280.65} \). The value will be in \( kJ/K \).
05

Find the total increase in entropy

The total increase in entropy is the sum of the entropy change during melting and heating. Therefore, \( \Delta S = \Delta S1 + \Delta S2 \)

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Most popular questions from this chapter

You're the environmental protection officer for a \(35 \%\) efficient nuclear power plant that produces 750 MW of electric power, situated on a river whose minimum flow rate is \(110 \mathrm{m}^{3} / \mathrm{s}\). State environmental regulations limit the rise in river temperature from your plant's cooling system to \(5^{\circ} \mathrm{C}\). Can you achieve this standard if you use river water for all your cooling, or will you need to install cooling towers that transfer some of your waste heat to the atmosphere?

The temperature of \(n\) moles of ideal gas is changed from \(T_{1}\) to \(T_{2}\) with pressure held constant. Show that the corresponding entropy change is \(\Delta S=n C_{p} \ln \left(T_{2} / T_{1}\right)\).

Find an expression for the entropy gain when hot and cold water are irreversibly mixed. A corresponding reversible process you can use to calculate this change is to bring each water sample slowly to their common final temperature \(T_{\mathrm{f}}\) and then mix them. Express your answer in terms of the initial temperatures \(T_{\mathrm{h}}\) and \(T_{\mathrm{c}} .\) Assume equal masses of hot and cold water, with constant specific heat \(c .\) What's the sign of your answer?

A refrigerator maintains an interior temperature of \(4^{\circ} \mathrm{C}\) while its exhaust temperature is \(30^{\circ} \mathrm{C} .\) The refrigerator's insulation is imperfect, and heat leaks in at the rate of 340 W. Assuming the refrigerator is reversible, at what rate must it consume electrical energy to maintain a constant \(4^{\circ} \mathrm{C}\) interior?

Could you cool the kitchen by leaving the refrigerator open? Explain.
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