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Chapter 5: Problem 15

If a typical metabolic intake is \(2,000 \mathrm{kcal}\) each day, approximately how much energy does this translate to for one day, in units of \(\mathrm{kWh}\) ? Compare this to a typical American household's electricity usage of \(30 \mathrm{kWh}\) in a day.

Short Answer

Expert verified
Answer: The daily 2,000 kcal metabolic intake represents approximately 7.76% of a typical American household's daily electricity usage.

Step by step solution

01

Convert kcal to Joules

To convert the metabolic intake from kcal to Joules, we need to know the conversion factor. 1 kcal is equal to 4184 Joules. So, we will multiply the intake value in kcal by the conversion factor: \(2000 \mathrm{kcal} \times 4184 \frac{\mathrm{J}}{\mathrm{kcal}}\)
02

Calculate the energy in Joules

Now we perform the multiplication: \(2000 \mathrm{kcal} \times 4184 \frac{\mathrm{J}}{\mathrm{kcal}} = 8376000 \mathrm{J}\) So, the energy intake translates to \(8,376,000 \mathrm{J}\).
03

Convert Joules to kWh

Now we need to convert the result from Joules to kilowatt-hours (kWh). To do this, we must divide by the conversion factor. There are \(3.6 \times 10^6 \mathrm{J}\) in 1 kWh. Therefore, we will divide the intake value in Joules by the conversion factor: \(\frac{8376000 \mathrm{J}}{3.6 \times 10^6 \frac{\mathrm{J}}{\mathrm{kWh}}}\)
04

Calculate the energy in kWh

Perform the division: \(\frac{8376000 \mathrm{J}}{3.6 \times 10^6 \frac{\mathrm{J}}{\mathrm{kWh}}} = 2.327 \mathrm{kWh}\) So, the daily metabolic intake translates to approximately \(2.327 \mathrm{kWh}\).
05

Compare energy intake to electricity usage

Now that we have the daily energy intake in kWh, we can compare it to the 30 kWh that a typical American household uses in a day. To do this, we can calculate the percentage of the household's daily electricity usage that the metabolic intake represents. \(\frac{2.327 \mathrm{kWh}}{30 \mathrm{kWh}} \times 100\)
06

Calculate the percentage

Perform the calculation: \(\frac{2.327 \mathrm{kWh}}{30 \mathrm{kWh}} \times 100 \approx 7.76\%\) So, the daily metabolic intake of \(2,000 \mathrm{kcal}\) translates to approximately \(2.327 \mathrm{kWh}\), which is about \(7.76\%\) of a typical American household's daily electricity usage (30 kWh).

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Most popular questions from this chapter

Gather up or compute conversion factors from the chapter to start your own conversion table (empty version below). Express kWh, cal, kcal, Btu, and Therms in terms of Joules.

Houses in the U.S. are equipped with circuit protection rated to 100 or 200 Amps, typically. If a 100 A house is operating at \(80 \%\) of its rated capacity, \(^{52}\) how much power is it consuming (at \(120 \mathrm{~V}\) )? If sustained for a month, how many kWh will show up on the bill? At \(\$ 0.15 / \mathrm{kWh}\), what is the cost?

At what wavelength, in microns \((\mu \mathrm{m})\), is the corresponding photon energy in eV the same number? A deliberately wrong example to illustrate would be if a \(2.6 \mu \mathrm{m}\) wavelength corresponded to \(2.6 \mathrm{eV}\) (it doesn't').

Just for fun, compute the energy associated with the mass of a tiny bit of shaving stubble having a mass of \(0.01 \mathrm{mg}^{41}\) using \(E=m c^{2}\). Make sure you use the correct units to put the result in Joules. The speed of light, \(c\), is approximately \(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\)

A typical textbook may have a mass of \(1 \mathrm{~kg}\), and thus a weight of about \(10 \mathrm{~N}\). How high could the textbook be lifted (against the force of gravity) by supplying one Joule of energy?
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