- #1
pastro
- 15
- 0
Hello,
I was wondering:
[tex]\frac{dQ}{dt}[/tex] = \sigma A T^{4}
for a perfect blackbody.
Also
Q = mc\DeltaT
If I take the time derivative of the above equation, set it equal to the power emitted by a blackbody, and solve the resulting differential equation for temperature, does that give me the temperature with which a blackbody radiator of a given mass and material cools in vacuum as a function of time?
Just curious...
Thanks!
I was wondering:
[tex]\frac{dQ}{dt}[/tex] = \sigma A T^{4}
for a perfect blackbody.
Also
Q = mc\DeltaT
If I take the time derivative of the above equation, set it equal to the power emitted by a blackbody, and solve the resulting differential equation for temperature, does that give me the temperature with which a blackbody radiator of a given mass and material cools in vacuum as a function of time?
Just curious...
Thanks!
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