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Chapter 6: Problem 94

One way to lose weight is to exercise! Walking briskly at 4.0 miles per hour for an hour consumes about \(400 \mathrm{kcal}\) of energy. How many hours would you have to walk at \(4.0\) miles per hour to lose one pound of body fat? One gram of body fat is equivalent to \(7.7 \mathrm{kcal}\) of energy. There are \(454 \mathrm{~g}\) in \(1 \mathrm{lb}\).

Short Answer

Expert verified
You would have to walk for approximately \(8.74 \mathrm{hours}\) at \(4.0 \mathrm{miles/hour}\) to lose one pound of body fat.

Step by step solution

01

Calculate kcal in one pound of body fat

We know that 1 gram of body fat is equivalent to 7.7 kcal of energy, and there are 454 grams in 1 lb. To find out how many kcal are in one pound of body fat, we will multiply the kcal per gram of fat by the number of grams in a pound. \[kcal_{1lb} = 7.7 \mathrm{kcal/g} \times 454 \mathrm{g}\]
02

Calculate kcal in one pound of body fat

Now, let's perform the calculation: \[kcal_{1lb} = 7.7 \mathrm{kcal/g} \times 454 \mathrm{g} = 3495.8 \mathrm{kcal}\] So there are approximately 3495.8 kcal in one pound of body fat.
03

Calculate hours required to lose one pound

Since walking at 4.0 miles per hour for one hour burns 400 kcal of energy, we can now divide the total kcal in one pound of body fat by the kcal burned per hour to find the number of hours required to lose one pound of body fat. \[hours = \frac{kcal_{1lb}}{kcal_{burned/hour}}\]
04

Calculate hours required to lose one pound

Now, let's perform the calculation using the values we found: \[hours = \frac{3495.8 \mathrm{kcal}}{400 \mathrm{kcal/hour}} \approx 8.74 \mathrm{hours}\] Therefore, you would have to walk for about 8.74 hours at 4.0 miles per hour to lose one pound of body fat.

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Most popular questions from this chapter

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) are gaseous fuels with enthalpies of combustion of \(-49.9 \mathrm{~kJ} / \mathrm{g}\) and \(-49.5 \mathrm{~kJ} / \mathrm{g}\), respectively. Compare the energy available from the combustion of a given volume of acetylene to the combustion energy from the same volume of butane at the same temperature and pressure.

A system releases \(125 \mathrm{~kJ}\) of heat while \(104 \mathrm{~kJ}\) of work is done on it. Calculate \(\Delta E\).

The bomb calorimeter in Exercise 108 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). A \(1.056-\mathrm{g}\) sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{~kJ} / \mathrm{g}\right)\) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

A sample consisting of \(22.7 \mathrm{~g}\) of a nongaseous, unstable compound \(\mathrm{X}\) is placed inside a metal cylinder with a radius of \(8.00 \mathrm{~cm}\), and a piston is carefully placed on the surface of the compound so that, for all practical purposes, the distance between the bottom of the cylinder and the piston is zero. (A hole in the piston allows trapped air to escape as the piston is placed on the compound; then this hole is plugged so that nothing inside the cylinder can escape.) The piston-and-cylinder apparatus is carefully placed in \(10.00 \mathrm{~kg}\) water at \(25.00^{\circ} \mathrm{C}\). The barometric pressure is 778 torr. When the compound spontaneously decomposes, the piston moves up, the temperature of the water reaches a maximum of \(29.52^{\circ} \mathrm{C}\), and then it gradually decreases as the water loses heat to the surrounding air. The distance between the piston and the bottom of the cylinder, at the maximum temperature, is \(59.8 \mathrm{~cm}\). Chemical analysis shows that the cylinder contains \(0.300 \mathrm{~mol}\) carbon dioxide, \(0.250\) mol liquid water, \(0.025\) mol oxygen gas, and an undetermined amount of a gaseous element \(\mathrm{A}\). It is known that the enthalpy change for the decomposition of \(X\), according to the reaction described above, is \(-1893\) \(\mathrm{kJ} / \mathrm{mol} \mathrm{X}\). The standard enthalpies of formation for gaseous carbon dioxide and liquid water are \(-393.5 \mathrm{~kJ} / \mathrm{mol}\) and \(-286 \mathrm{~kJ} / \mathrm{mol}\), respectively. The heat capacity for water is \(4.184 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\). The conversion factor between \(\mathrm{L} \cdot \mathrm{atm}\) and \(\mathrm{J}\) can be determined from the two values for the gas constant \(R\), namely, \(0.08206 \mathrm{~L}\). \(\mathrm{atm} / \mathrm{K} \cdot \mathrm{mol}\) and \(8.3145 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). The vapor pressure of water at \(29.5^{\circ} \mathrm{C}\) is 31 torr. Assume that the heat capacity of the pistonand-cylinder apparatus is negligible and that the piston has negligible mass. Given the preceding information, determine a. The formula for \(\mathrm{X}\). b. The pressure-volume work (in \(\mathrm{kJ}\) ) for the decomposition of the \(22.7-\mathrm{g}\) sample of \(\mathrm{X}\). c. The molar change in internal energy for the decomposition of \(X\) and the approximate standard enthalpy of formation for \(X\).

The sun supplies energy at a rate of about \(1.0\) kilowatt per square meter of surface area ( 1 watt \(=1 \mathrm{~J} / \mathrm{s}\) ). The plants in an agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare ( \(1 \mathrm{ha}=10,000 \mathrm{~m}^{2}\) ). Assuming that sucrose is produced by the reaction \(12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g)\) \(\Delta H=5640 \mathrm{~kJ}\) calculate the percentage of sunlight used to produce the sucrosethat is, determine the efficiency of photosynthesis.
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