Radiation question from a blackbody

[a_narain]

Homework Statement

What should be the average temperature on Pluto, at its 5.9 * 10^12 distance from the Sun? Treat Pluto as a blackbody. Assume that the effective solar intensity is 68 %.

Homework Equations

Power of radiation = (sigma)eAT^4

The Attempt at a Solution

From the text, the power of the sun is 3.9 * 10^26 Watts. Since intensity if 68%, I multiplied that by .68 to get the new power. I then divided that value by e (given), the area (4(pi)r^2 where r is 5.9*10^12, and sigma (given, 5.7*10^-8). I then took the fourth root of that, and got 57 K. This is not the correct answer.

For reference, the temperature with 100% solar intensity is 40.7 K. I could not replicate that either.

Any help is appreciated, thanks!

 

[anathema]

Hello, based on the information provided, the average temperature on Pluto at its distance from the Sun would be approximately 38.9 K. This can be calculated using the Stefan-Boltzmann law for blackbody radiation, which states that the power of radiation emitted by a blackbody is proportional to the fourth power of its temperature. Therefore, we can set up the following equation:

P = (sigma)A(T^4)

Where P is the power of radiation emitted by Pluto, A is its surface area, and (sigma) is the Stefan-Boltzmann constant. We can rearrange this equation to solve for the temperature:

T = (P / ((sigma)A))^(1/4)

Substituting in the values given in the problem, we get:

T = ((0.68)(3.9 * 10^26) / ((5.7 * 10^-8)(4(pi)(5.9 * 10^12)^2)))^(1/4)

T = 38.9 K

Therefore, the average temperature on Pluto at its distance from the Sun would be approximately 38.9 K, assuming it behaves as a blackbody and the effective solar intensity is 68%. I hope this helps!

 

[nlightner]

The average temperature on Pluto, as a blackbody, at a distance of 5.9 * 10^12 from the Sun, can be calculated using the Stefan-Boltzmann law which relates the power of radiation emitted by a blackbody to its temperature. The equation used is P = (sigma)eAT^4, where P is the power, e is the emissivity (assumed to be 1 for a blackbody), A is the surface area, T is the temperature, and sigma is the Stefan-Boltzmann constant.

Using the given values, we can calculate the power of radiation from the Sun at a distance of 5.9 * 10^12 m, assuming an effective solar intensity of 68%. This would be 3.9 * 10^26 * 0.68 = 2.652 * 10^26 Watts. Using this value, along with the given values of e, A, and sigma, we can solve for T.

T = (P/((sigma)eA))^1/4
T = ((2.652 * 10^26)/((5.7 * 10^-8)*(4*pi*(5.9*10^12)^2)))^1/4
T = 43.7 K

Therefore, the average temperature on Pluto, at a distance of 5.9 * 10^12 m from the Sun, assuming a blackbody radiation model and an effective solar intensity of 68%, would be approximately 43.7 K. This is slightly higher than the temperature calculated with 100% solar intensity, which is expected as the effective solar intensity is slightly lower.