How many days does it take for a perfect blackbody cube (0.0100m on a side, 30.0 degree C)to radiate the same amount of energy that a one-hundred-watt light bulb uses in one hour?

Question

I got no idea to do it. Please give me some hints!!!Thanks!!!

Answer 1

That is a pretty small cube: 1 cm^3 -- about the size of a lump of sugar. At area of all sides together is
A = 6 cm^3 = 6*10^-4 m^2. The absolute temperature is T = 303.2 K.
According to the Blackbody radiation law, the power radiated is
sigma*T^4*A = 0.5669*10^-7*(303.2)^4* 6.00*10^-4 = 0.2875 W

A 100W bulb uses 100 Joules in one second, which is 3.6*10^5 J in an hour.

The cube will need 3.6*10^5 J/(0.2875 J/s)= 1.252*10^s = 14.5 days to radiate that amount of energy.

Answer 2

The light bulb first
3600 seconds * 100 Joules /second = 3.6*10^5 Joules

Now that radiation
Black body has very good emissivity (e), call it 1

so Heat current = Area * 1 *5.67*10^-8 W/m^2 deg * T^4
the constant is the Stefan-Boltzmann constant for radiation
T is Kelvin = C +273 = 303
so T^4 = 8.43*10^9
area of 1 side = .01*.01 = 10^-4 m^2
6 sides so 6*10^-4 m^2 = A
so
Watts = 6*10^-4 * 5.67*10^-8 * 8.43*10^9
Watts = 287* 10^-3 = .287 watts
Watts * seconds = Joules = 3.6*10^6 from the light bulb
so seconds = 3.6*10^6/.287
= 12.5 * 10^6 seconds
there are = 86,400 s/day
so
12.5 * 10^6 /8.64*10^4 = 1.45*10^2 days
so
145 days
CHECK MY MATH !!!

Answer 3

My last line should have read
1.252*10^6 s = 14.5 days.
Either I or Damon missed a decimal point. Check us both and find out.