Asked by jean
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
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
Answers
Answered by
BECKY
1000 = (5.67*10^8)(1m^2)(T^4)
Then solve for T :)
Then solve for T :)
Answered by
BECKY
OH! Then subtract 273 from answer to change it to celcius.
Answered by
BEREKET
Estimate the temperature of a blacktop road on a sunny day. Assume the asphalt
is a perfect blackbody
is a perfect blackbody
Answered by
tom
you mean 5.67*10^-8
There are no AI answers yet. The ability to request AI answers is coming soon!