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Chapter 17: Problem 114

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 \(\Delta S_{\text {surr }}\) 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 \(75.4 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\).

Short Answer

Expert verified
The change in entropy of the surroundings as the hot water cools down to room temperature is approximately \(-869.10 \, J/K\).

Step by step solution

01

Determine the mass of water

First, convert the volume of water from L to cm³: 1.00 L = 1000 cm³. To find the mass of the water, multiply its volume by its density: Mass = Volume × Density Mass = 1000 cm³ × 1.00 g/cm³ = 1000 g.
02

Calculate the number of moles of water

To find the number of moles of water, use the molar mass of water (18 g/mol): n = Mass / Molar Mass n = 1000 g / 18 g/mol ≈ 55.56 mol.
03

Calculate the heat lost by the water

To find the heat transfer (q) from the water to the surroundings, we can use the formula: q = n × C × ΔT where n is the number of moles, C is the heat capacity (75.4 J/mol·K for water), and ΔT is the temperature change from 90°C to 25°C. ΔT = T_final - T_initial ΔT = 25°C - 90°C = -65°C Now we can plug in the values in the heat transfer formula: q = 55.56 mol × 75.4 J/mol·K × (-65°C) ≈ -259130 J. Since entropy is a state function, we can focus on the absolute values of temperature and heat transfer to find the change in entropy of the surroundings.
04

Calculate the change in entropy for the surroundings

The change in entropy (ΔS) of the surroundings can be determined by dividing the transferred heat by the ambient temperature: ΔS_surr = q / T Since q is negative (heat lost by the system) and T is always positive, ΔS_surr would be negative as well. We convert T from Celsius to Kelvin: T = 25°C + 273.15 = 298.15 K Now, calculate ΔS_surr: ΔS_surr = -259130 J / 298.15 K ≈ -869.10 J/K. The change in entropy of the surroundings is -869.10 J/K.

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Most popular questions from this chapter

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ}\) ? a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

For the reaction at \(298 \mathrm{~K}\), $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ 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 \(\Delta H^{\circ}\) 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?

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

Given the values of \(\Delta H\) and \(\Delta S\), which of the following changes will be spontaneous at constant \(T\) and \(P\) ? a. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

For rubidium \(\Delta H_{\mathrm{vap}}^{\circ}=69.0 \mathrm{~kJ} / \mathrm{mol}\) at \(686^{\circ} \mathrm{C}\), its boiling point. Calculate \(\Delta S^{\circ}, q, w\), and \(\Delta E\) for the vaporization of \(1.00 \mathrm{~mole}\) of rubidium at \(686^{\circ} \mathrm{C}\) and \(1.00\) atm pressure.
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