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Chapter 6: Problem 95

It has been determined that the body can generate \(5500 \mathrm{~kJ}\) of energy during one hour of strenuous exercise. Perspiration is the body's mechanism for eliminating this heat. What mass of water would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise? (The heat of vaporization of water is \(40.6 \mathrm{~kJ} / \mathrm{mol}\).)

Short Answer

Expert verified
The mass of water that would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise is approximately \(4878.53 \mathrm{~g}\) or \(4.88 \mathrm{~kg}\).

Step by step solution

01

Calculate the total energy generated in 2 hours of exercise

First, we need to calculate the total energy generated during 2 hours of exercise. We are given that the body can generate \(5500 \mathrm{~kJ}\) of energy during one hour of exercise. To find the total energy in 2 hours, we simply multiply the energy generated in one hour by 2: Total energy generated = Energy generated in one hour × Time (in hours) = \(5500 \mathrm{~kJ/hour} \times 2 \mathrm{~hours} = 11000 \mathrm{~kJ}\)
02

Calculate the number of moles of water evaporated

We are given the heat of vaporization of water, which is \(40.6 \mathrm{~kJ/mol}\). To find the number of moles of water needed to be evaporated, we can divide the total energy generated by the heat of vaporization of water: Number of moles = Total energy generated / Heat of vaporization of water = \(11000 \mathrm{~kJ} / 40.6 \mathrm{~kJ/mol} = 270.94 \mathrm{~mol}\)
03

Calculate the mass of water evaporated

Now that we have the number of moles of water that needs to be evaporated, we can convert it into mass using the molar mass of water. The molar mass of water is \(18.015 \mathrm{~g/mol}\). Mass of water evaporated = Number of moles × Molar mass of water = \(270.94 \mathrm{~mol} \times 18.015 \mathrm{~g/mol} = 4878.53 \mathrm{~g}\) Thus, the mass of water that would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise is approximately \(4878.53 \mathrm{~g}\) or \(4.88 \mathrm{~kg}\).

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Most popular questions from this chapter

At \(298 \mathrm{~K}\), the standard enthalpies of formation for \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are \(227 \mathrm{~kJ} / \mathrm{mol}\) and \(49 \mathrm{~kJ} / \mathrm{mol}\), respectively. a. Calculate \(\Delta H^{\circ}\) for $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) $$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

On Easter Sunday, April 3,1983, nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: \(2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \% \mathrm{HNO}_{3}\) by mass with a density of \(1.42 \mathrm{~g} / \mathrm{cm}^{3} .\) What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\left.\mathrm{NaNO}_{3}(a q)=-467 \mathrm{~kJ} / \mathrm{mol}\right)\) b. According to The Denver Post for April 4,1983 , authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\), what was their major concern?

A \(30.0\) -g sample of water at \(280 . \mathrm{K}\) is mixed with \(50.0 \mathrm{~g}\) water at \(330 . \mathrm{K}\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Why is it a good idea to rinse your thermos bottle with hot water before filling it with hot coffee?

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.
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