Posted by **jean** on Thursday, April 26, 2007 at 12:13am.

At high noon, the Sun delivers 1000W to each square meter of a blacktop road. If the hot asphalt loses energy only by radiation, what is its equilibrium temperature?

so I know that the formula to find the RATE OF RADIATION= the constant (5.6696 x 10^-8 W/m^2*K^4)*A*e*(T^4)

but it doesn't seem to help me much for this problem

Just set the 1000 W/m^2 equal to the radiated energy given by the Stefan-Botlzmann equation. Then solve for T. I don't know what the "e" is doing in your equation

sigma *Area * T^4 = 1000 W

sinma = (5.6696 x 10^-8 W/m^2*K^4)

the thing is, there's no area given, and i'm not sure how to find it

The area is one square meter. The "sigma" constant gives energy radiated per squate meter, and the the 1000 "solar constant" number is also per square meter

- physics -
**BECKY**, Saturday, October 15, 2011 at 1:58am
1000 = (5.67*10^8)(1m^2)(T^4)

Then solve for T :)

- physics -
**BECKY**, Saturday, October 15, 2011 at 2:00am
OH! Then subtract 273 from answer to change it to celcius.

- physics -
**BEREKET**, Thursday, September 3, 2015 at 5:18am
Estimate the temperature of a blacktop road on a sunny day. Assume the asphalt

is a perfect blackbody

- physics -
**tom**, Thursday, April 14, 2016 at 1:13am
you mean 5.67*10^-8

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