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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 -

    1000 = (5.67*10^8)(1m^2)(T^4)
    Then solve for T :)

  • physics -

    OH! Then subtract 273 from answer to change it to celcius.

  • physics -

    Estimate the temperature of a blacktop road on a sunny day. Assume the asphalt
    is a perfect blackbody

  • physics -

    you mean 5.67*10^-8

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