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Chapter 11: Problem 43

For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water? (The heat of vaporization of water in this temperature range is \(2.4 \mathrm{~kJ} / \mathrm{g}\). The specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} .)\)

Short Answer

Expert verified
1. Calculate \(Q_\mathrm{evap} = 60 \mathrm{~g} \cdot 2.4 \mathrm{~kJ/g} = 144 \mathrm{~kJ}\) 2. Convert \(Q_\mathrm{evap}\) to joules: \(144 \mathrm{~kJ} \cdot 1000 = 144,000 \mathrm{~J}\) 3. Calculate \(\Delta T = 35^\circ \mathrm{C} - 20^\circ \mathrm{C} = 15^\circ \mathrm{C}\) 4. Use the mass formula to find \(m\): \(m = \frac{144,000 \mathrm{~J}}{4.18 \mathrm{~J/gK} \cdot 15^\circ \mathrm{C}} \approx 2290\,\mathrm{g}\) Thus, approximately \(2290 \mathrm{~g}\) of water can be cooled from \(35^\circ \mathrm{C}\) to \(20^\circ \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water.

Step by step solution

01

Calculate the heat absorbed by the evaporated water

To find the total heat absorbed by the evaporated water, we need to multiply the mass of the evaporated water (60 g) by its heat of vaporization (\(2.4 \mathrm{~kJ} / \mathrm{g}\)): $$ Q_\mathrm{evap} = m_\mathrm{evap} \cdot L_\mathrm{v} $$ where \(Q_\mathrm{evap}\) is the heat absorbed by the evaporated water, \(m_\mathrm{evap}=60\,\mathrm{g}\) is the mass of the evaporated water, and \(L_\mathrm{v}=2.4\,\mathrm{kJ/g}\) is the heat of vaporization of water.
02

Convert the heat absorbed into joules

Since the specific heat of water is given in joules, it is necessary to convert the heat absorbed by the evaporated water from kilojoules to joules: $$ Q_\mathrm{evap(J)} = Q_\mathrm{evap(kJ)} \cdot 1000 $$
03

Calculate the mass of cooled water

Now, we'll use the heat absorbed by the evaporated water to find out how many grams of water can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\). To do this, we use the formula: $$ Q = m \cdot c \cdot \Delta T $$ where \(Q\) is the heat absorbed, \(m\) is the mass of cooled water, \(c=4.18 \,\mathrm{J/gK}\) is the specific heat of water, and \(\Delta T = 35^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C}\) is the change in temperature. Rearrange the formula to find the mass of cooled water: $$ m = \frac{Q}{c \cdot \Delta T} $$
04

Combine the steps and solve for m

From steps 1-3, we have all the necessary information to solve for \(m\). Plug the values into the formula from step 3 and solve for the mass of cooled water: 1. Calculate \(Q_\mathrm{evap} = 60 \mathrm{~g} \cdot 2.4 \mathrm{~kJ/g}\) 2. Convert \(Q_\mathrm{evap}\) to joules 3. Calculate \(\Delta T = 35^\circ \mathrm{C} - 20^\circ \mathrm{C}\) 4. Use the mass formula to find \(m\) After solving for \(m\), we will have the mass of water that can be cooled from \(35^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\) by the evaporation of \(60 \mathrm{~g}\) of water.

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Most popular questions from this chapter

Appendix B lists the vapor pressure of water at various external pressures. (a) Plot the data in Appendix B, vapor pressure (torr) versus temperature \(\left({ }^{\circ} \mathrm{C}\right) .\) From your plot, estimate the vapor pressure of water at body temperature, \(37^{\circ} \mathrm{C}\). (b) Explain the significance of the data point at 760.0 torr, \(100^{\circ} \mathrm{C}\) (c) A city at an altitude of \(5000 \mathrm{ft}\) above sea level has a barometric pressure of 633 torr. To what temperature would you have to heat water to boil it in this city? (d) A city at an altitude of \(500 \mathrm{ft}\) below sea level would have a barometric pressure of 774 torr. To what temperature would you have to heat water to boil it in this city? (e) For the two cities in parts \((\mathrm{c})\) and \((\mathrm{d}),\) compare the average kinetic energies of the water molecules at their boiling points. Are the kinetic energies the same or different? Explain.

The generic structural formula for a 1 -alkyl-3-methylimidazolium cation is where \(\mathrm{R}\) is a \(-\mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{n} \mathrm{CH}_{3}\) alkyl group. The melting points of the salts that form between the 1 -alkyl-3-methylimidazolium cation and the \(\mathrm{PF}_{6}^{-}\) anion are as follows: \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=60^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{m.p.}=40^{\circ} \mathrm{C}\right)\) \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=10^{\circ} \mathrm{C}\right)\) and \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=-61^{\circ} \mathrm{C}\right) . \mathrm{Why}\) does the melting point decrease as the length of alkyl group increases?

Rationalize the difference in boiling points in each pair: (a) \(\mathrm{HF}\left(20^{\circ} \mathrm{C}\right)\) and \(\mathrm{HCl}\left(-85^{\circ} \mathrm{C}\right),(\mathbf{b}) \mathrm{CHCl}_{3}\left(61{ }^{\circ} \mathrm{C}\right)\) and \(\mathrm{CHBr}_{3}\) \(\left(150^{\circ} \mathrm{C}\right),(\mathrm{c}) \mathrm{Br}_{2}\left(59^{\circ} \mathrm{C}\right)\) and \(\mathrm{ICl}\left(97^{\circ} \mathrm{C}\right)\)

The phase diagram of a hypothetical substance is (a) Estimate the normal boiling point and freezing point of the substance. (b) What is the physical state of the substance under the following conditions: (i) \(T=150 \mathrm{~K}, P=0.2 \mathrm{~atm}\) (ii) \(T=100 \mathrm{~K}, P=0.8 \mathrm{~atm},(\mathrm{iii}) T=300 \mathrm{~K}, P=1.0 \mathrm{~atm} ?\) (c) What is the triple point of the substance? [Section 11.6\(]\)

If \(42.0 \mathrm{~kJ}\) of heat is added to a 32.0 -g sample of liquid methane under 1 atm of pressure at a temperature of \(-170{ }^{\circ} \mathrm{C},\) what are the final state and temperature of the methane once the system equilibrates? Assume no heat is lost to the surroundings. The normal boiling point of methane is \(-161.5^{\circ} \mathrm{C}\). The specific heats of liquid and gaseous methane are 3.48 and \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K},\) respectively. [Section 11.4\(]\)
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