Statistical Physics: November 15, 2012
Solutions for the Homework 8
Problem 7.52:
(a) Estimate (roughly) the total power radiated by your body, neglecting any energy that is
returned by your clothes and environment. (Whatever the color of your skin, its emissivity at
infrared wavelengths is quite close to 1; almost any nonmetal is a near-perfect blackbody at these
wavelengths.)
Solution: Let’s assume that the area of human body is almost 1 m2 , temperature is 310 K and
the emissivity is equal to 1. Then,
power = σeAT 4 = 523.673 J/s.
(1)
(b) Compare the total energy radiated by your body in one day (expressed in kilocalories) to
the energy in the food you eat. Why is there such a large discrepancy?
Solution: One day is 86, 400 seconds and 1 kcal is 4184 J. So, the total energy radiated by
a human body is almost 10, 813.9 kcal. This is 4 times daily requirement of energy from food
and it’s ridiculous. The reason is that we can also get the radiated energy from the other things
neighboring me and it’s also similar order of that.
(c) The sun has a mass of 2 × 1030 kg and radiates energy at a rate of 3.9 × 1026 watts. Which
puts out more power per units massthe sun or your body?
Solution: Let’s assume that the mass of someone is 80kg. Then, put the value and compare
the energy and mass ratio. Then human gives more energy with respect to the unit mass and
the value is 33, 568.8 times higher than the sun case. Because of the reaction of the sun power
is more slower than the chemical reaction of human body.
Problem 7.54: The sun is the only star whose size we can easily measure directly; astronomers
therefore estimate the sizes of other stars using Stefan’s law.
(a) The spectrum of Sirius A, plotted as a function of energy, peaks at a photon energy of 2.4eV ,
while Sirius A is approximately 24 times as luminous as the sun. How does the radius of Sirius
A compare to the sun’s radius?
Solution: Find temperature using the photon energy peak, max = 2.82kT . Then, substitute
the temperature to the equation, that is
4πR2 σT 4 = luminosity ratio with respect to the sun.
(2)
Now, solve this equation to R and then, R = 1.17503 × 109 m which is 1.68947 times the radius
of the sun.
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Statistical Physics: November 15, 2012
(b) Sirius B, the companion of Sirius A (see Figure 7.12), is only 3% as luminous as the sun. Its
spectrum, plotted as a function of energy, peaks at about 7eV . How does its radius compare to
that of the sun?
Solution: Similarly, 4.88348 × 106 m which is 0.00702154 times the radius of the sun.
(c) The spectrum of the star Betelgeuse, plotted as a function of energy, peaks at a photon
energy of 0.8eV , while Betelgeuse is approximately 10, 000 times as luminous as the sun. How
does the radius of Betelgeuse compare to the sun’s radius? Why is Betelgeuse called a “red
supergiant ”?
Solution: Similarly, 2.15866 × 1011 m which is 310.376 times the radius of the sun. When
you calculate the wavelength of maximum energy, it will red and so this star is called a “red
supergiant ”.
Problem 7.56: The planet Venus is different from the earth in several respects. First, it is only 70%
as far from the sun. Second, its thick clouds reflect 77% of all incident sunlight. Finally, its
atmosphere is much more opaque to infrared light.
(a) Calculate the solar constant at the location of Venus, and estimate what the average surface
temperature of Venus would be if it had no atmosphere and did not reflect any sunlight.
Solution: Use the equation 7.102, that the statement of equilibrium which means that the
receiving energy should same as the radiating energy. The only change thing is the distance
from the sun and it changes the solar constant as a factor 1/(0.7)2 :
1
(solar constant) ×
× πR2 = 4πR2 σT 4 .
(3)
(0.7)2
Then the temperature is 333.206 K.
(b) Estimate the surface temperature again, taking the reflectivity of the clouds into account.
Solution: Multiply the absorption rate to the solar constant using the reflexivity R:
1
πR2 = 4πR2 σT 4 .
(solar constant)(1 − R)
(0.7)2
(4)
Then the temperature is 230.752 K.
(c) The opaqueness of Venus’s atmosphere at infrared wavelengths is roughly 70 times that of
earth’s atmosphere. You can therefore model the atmosphere of Venus as 70 successive “blankets
”of the type considered in the text, with each blanket at a different equilibrium temperature.
Use this model to estimate the surface temperature of Venus. (Hint: The temperature of the
top layer is what you found in part (b). The next layer down is warmer by a factor of 21/4 . The
next layer down is warmer by a smaller factor. Keep working your way down until you see the
pattern.)
Solution: Consider the figure 7.25. There is only one atmosphere but we need 70 atmospheres
like blankets. So, change the ground as an atmosphere and the ground is below of that atmosphere. Since the second atmosphere radiate the energy, 2 times receiving sunlight energy to
upward, then also downward. To make the ground is thermal equilibrium, ground should radiate 3 times energy to upward. Repeat this process to the 70th blanket. Then the ground will
radiate 71 times energy to upward. Using this, multiply this factor to the total power. Then,
the temperature is 669.822 K.
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