How Do Meters, Kelvins, and Watts Relate in This Equation?

[tony873004]

Units -- Watts, meters & Kelvin

I have an equation:
$$L = \sigma T^{4} * 4 \pi R^{2}$$

Which I fill with values and units
$$L=5.67*10^{-8} * (5860K)^{4} * 4 \pi * (6.96*10^{8}m)^{2}$$

and I get an answer:

$$4.07*10^{26} K^{4} m^{2}$$

The correct answer is:

$$4.07*10^{26} watts$$

I don't understand the relationship between meters, Kelvins, and Watts. How do I properly combine or cancel the units to arrive at Watts?

 

[Integral]

Investigate the units on your $\sigma$

 

[tony873004]

Investigate the units on your $\sigma$

I was thinking the answer might lie there, but that might be a unitless constant. I think there is a Kelvin meter relationship that yields watts.

 

[chroot]

Sigma is the Stefan-Boltzmann constant, and is certainly not unitless.

http://scienceworld.wolfram.com/physics/Stefan-BoltzmannConstant.html

- Warren

 

[tony873004]

Thank you. I get it now. You're both right. It does have units