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Chapter 1: Problem 24

Verify the total solar power output of \(4 \times 10^{26} \mathrm{~W}\) based on its surface temperature of \(5,800 \mathrm{~K}\) and radius of \(7 \times 10^{8} \mathrm{~m}\), using Eq. \(1.9 .\)

Short Answer

Expert verified
Question: Verify the total solar power output of the Sun given its surface temperature of 5,800 K and radius of \(7 \times 10^8 \mathrm{~m}\), using the Stefan-Boltzmann Law. The given total solar power output is \(4 \times 10^{26} \mathrm{~W}\). Answer: The calculated theoretical power output using the Stefan-Boltzmann Law is approximately \(3.99 \times 10^{26} \mathrm{~W}\), which is very close to the given value of \(4 \times 10^{26} \mathrm{~W}\). This verifies the total solar power output of the Sun based on its surface temperature and radius.

Step by step solution

01

1. Write down the given data

Given: Total solar power output (L) = \(4 \times 10^{26} \mathrm{~W}\) Surface temperature (T) = \(5,800 \mathrm{~K}\) Radius (R) = \(7 \times 10^{8} \mathrm{~m}\)
02

2. Write down the Stefan-Boltzmann Law (Equation 1.9)

The Stefan-Boltzmann Law is given by: \(L = 4 \pi R^2 \sigma T^4\) Where \(L\) is the luminosity (total power output) of the star, \(R\) is the radius of the star, \(T\) is the surface temperature of the star, and \(\sigma\) is the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8} \mathrm{~W~m^{-2}~K^{-4}}\).
03

3. Plug in the given data and solve for L

Using the given values of \(R\) and \(T\), along with the Stefan-Boltzmann constant, we can now calculate the theoretical power output (L): \(L = 4 \pi (7 \times 10^{8} \mathrm{~m})^2 \cdot (5.67 \times 10^{-8} \mathrm{~W~m^{-2}~K^{-4}}) \cdot (5,800 \mathrm{~K})^4\)
04

4. Calculate the theoretical power output

Now, compute the value of L: \(L = 4 \pi (7 \times 10^8)^2 \cdot 5.67 \times 10^{-8} \cdot 5,800^4\) \(L \approx 3.99 \times 10^{26} \mathrm{~W}\)
05

5. Compare the theoretical power output with the given data

The calculated theoretical power output is approximately \(3.99 \times 10^{26} \mathrm{~W}\), which is very close to the given value of \(4 \times 10^{26} \mathrm{~W}\). This verifies the total solar power output of the Sun based on its surface temperature and radius, using Eq. 1.9 (Stefan-Boltzmann Law).

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Most popular questions from this chapter

Your skin temperature is about \(308 \mathrm{~K}\), and the walls in a typical room are about \(295 \mathrm{~K}\). If you have about \(1 \mathrm{~m}^{2}\) of outward-facing surface area, how much power do you radiate as infrared radiation, in Watts? Compare this to the typical metabolic rate of \(100 \mathrm{~W}\).

In the spirit of outlandish extrapolations, if we carry forward a \(2.3 \%\) growth rate \((10 \times\) per century \()\), how long would it take to go from our current \(18 \mathrm{TW}\left(18 \times 10^{12} \mathrm{~W}\right)\) consumption to annihilating an entire earth-mass planet every year, converting its mass into pure energy using \(E=m c^{2} ?\) Things to know: Earth's mass is \(6 \times 10^{24} \mathrm{~kg} ; c=3 \times 10^{8} \mathrm{~m} / \mathrm{s} ;\) the result is in Joules, and one Watt is one Joule per second.

Using Eq. \(1.5\) and showing work, what annual growth rate, in percent, leads to the mathematically convenient factor-of-ten growth every century?

A one-liter jar would hold about \(10^{16}\) bacteria. Based on the number of doubling times in our 24 -hour experiment, show by calculation that our setup was woefully unrealistic: that even if we started with a single bacterium, we would have far more than \(10^{16}\) bacteria after 24 hours if doubling every 10 minutes.

A more dramatic, if entirely unrealistic, version of the bacteria-jar story is having the population double every minute. Again, we start the jar with the right amount of bacteria so that the jar will be full 24 hours later, at midnight. At what time is the jar half full now?
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