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Chapter 17: Problem 124

You have a 1.00 -L sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\mathrm{sur}}\) for this process. Assume the density of water is 1.00 $\mathrm{g} / \mathrm{cm}^{3}$ over this temperature range, and the heat capacity of water is constant over this temperature range and equal to 75.4 $\mathrm{J} / \mathrm{K} \cdot$ mol.

Short Answer

Expert verified
The change in entropy of the surroundings, \(\Delta S_{\text{sur}}\), for the process of hot water cooling to room temperature is 871.6 J/K.

Step by step solution

01

Find the number of moles of water

First, you need to find the number of moles of water in the 1.00 L sample. Use the density of water, 1.00 g/cm³, to do this. Since 1 L = 1000 cm³, you have: Mass of water = Volume × Density = 1000 cm³ × 1.00 g/cm³ = 1000 g Now, find the number of moles of water, knowing that the molar mass of water is 18.02 g/mol: Number of moles = Mass / Molar mass = 1000 g / 18.02 g/mol = 55.49 mol
02

Calculate the heat exchanged

Now, calculate the amount of heat released by the water as it cools down from 90.0°C to 25.0°C. Use the heat capacity of water and the temperature difference: \(q = n \times C \times \Delta T\) \(q = 55.49 \mathrm{mol} \times 75.4 \mathrm{\frac{J}{K \cdot mol}} \times (25.0 ^\circ \mathrm{C} - 90.0 ^\circ \mathrm{C})\) \(q = 55.49 \mathrm{mol} \times 75.4 \mathrm{\frac{J}{K \cdot mol}} \times -65.0 \mathrm{K}\) \(q = -259910.3 \mathrm{J}\) The heat exchanged (q) is negative because it was released to the surroundings by the water sample.
03

Calculate the change in entropy of the surroundings

Finally, calculate the change in entropy of the surroundings with the formula: \(\Delta S_{\text{sur}} = -\frac{q}{T}\) Since the temperature of the room remains constant at 25.0°C, you can use that in the formula, but you need to convert it to Kelvin first: 25.0°C + 273.15 = 298.15 K \(\Delta S_{\text{sur}} = -\frac{-259910.3 \mathrm{J}}{298.15 \mathrm{K}}\) \(\Delta S_{\text{sur}} = 871.6 \mathrm{\frac{J}{K}}\) The change in entropy of the surroundings, \(\Delta S_{\text{sur}}\), is 871.6 J/K.

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Most popular questions from this chapter

At \(100 .^{\circ} \mathrm{C}\) and $1.00 \mathrm{atm}, \Delta H^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol}\( for the vaporization of water. Estimate \)\Delta G^{\circ}$ for the vaporization of water at \(90 .^{\circ} \mathrm{C}\) and \(110 .^{\circ} \mathrm{C}\) . Assume $\Delta H^{\circ}\( and \)\Delta S^{\circ}\( at \)100 .^{\circ} \mathrm{C}$ and 1.00 \(\mathrm{atm}\) do not depend on temperature.

For the reaction at 298 K, $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are $-58.03 \mathrm{kJ}\( and \)-176.6 \mathrm{J} / \mathrm{K}$ , respectively. What is the value of \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) ? Assuming that \(\Delta H^{p}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

Consider the following equilibrium constant versus temperature data for some reaction: $$\begin{array}{ll} {\boldsymbol{T}\left(^{\circ} \mathbf{C}\right)} & \quad {\text { K }} \\ \hline {109} & {2.54 \times 10^{4}} \\ {225} & {5.04 \times 10^{2}} \\ {303} & {6.33 \times 10^{1}} \\ {412} & {2.25 \times 10^{-1}} \\\ {539} & {3.03 \times 10^{-3}}\end{array}$$ Predict the signs for \(\Delta G^{\circ}, \Delta H^{\circ},\) and $\Delta S^{\circ}$ for this reaction at \(25^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and $\Delta S_{\mathrm{univ}}\( for the following process: \)\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)$

Choose the substance with the larger positional probability in each case. a. 1 mole of \(\mathrm{H}_{2}\) (at \(\mathrm{STP} )\) or 1 mole of $\mathrm{H}_{2}\left(\text { at } 100^{\circ} \mathrm{C}, 0.5 \mathrm{atm}\right)$ b. 1 mole of \(\mathrm{N}_{2}(\text { at } \mathrm{STP})\) or 1 mole of \(\mathrm{N}_{2}(\text { at } 100 \mathrm{K}, 2.0 \mathrm{atm})\) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(0^{\circ} \mathrm{C} )\) or 1 \(\mathrm{mole}\) of $\mathrm{H}_{2} \mathrm{O}(l)\left(\mathrm{at} 20^{\circ} \mathrm{C}\right)$
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