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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

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?

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.

In the more dramatic bacteria-jar scenario in which doubling happens every minute and reaches single-jar capacity at midnight, at what time will the colony have to cease expansion if an explorer finds three more equivalent jars in which they are allowed to expand without interruption/delay?

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 \(100 \mathrm{~W}\).

Verify the claim in the text that the town of 100 residents in 1900 reaches approximately 100,000 in the year 2000 if the doubling time is 10 years.
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