I How can a blackbody get hotter?

[UMath1]

TL;DR Summary

If rate of emission = rate of absorption for a blackbody, how can a blackbody undergo temperature change through radiation?

If rate of emission = rate of absorption for a blackbody, how can a blackbody undergo temperature change through radiation? How come black cars get hotter on a sunny day?

 

[Dale]

If rate of emission = rate of absorption for a blackbody, how can a blackbody undergo temperature change through radiation?

It wouldn’t.

How come black cars get hotter on a sunny day?

Because the rate of emission is lower than the rate of absorption.

You appear to be confusion the rate of emission with the emissivity.

 

[UMath1]

It wouldn’t.

Because the rate of emission is lower than the rate of absorption.

So a black car can't be modeled as a blackboday? What about a vantablack surface? These are used for solar energy generation and also get hotter?

 

[Dale]

So a black car can't be modeled as a blackboday? What about a vantablack surface? These are used for solar energy generation and also get hotter?

Both a black car and a vantablack surface can be modeled as a black body. You are misunderstanding how a black body works. Emissivity is not the same as the rate of emission

 

[UMath1]

Both a black car and a vantablack surface can be modeled as a black body. You are misunderstanding how a black body works. Emissivity is not the same as the rate of emission

I'm referencing this equation from Fundamentals of Heat and Mass Transfer. It suggests that net radiation to a blackbody is zero because Ebi=Ji and ei=1.

 

[Nugatory]

If rate of emission = rate of absorption...

That’s the equilibrium condition. If you surround your blackbody with hotter objects it will not be in equilibrium and the absorption rate will be greater than emission rate. As it heats up the emission rate will increase until it matches the absorption rate and equilibrium is reached at a higher temperature.

 

[UMath1]

That’s the equilibrium condition. If you surround your blackbody with hotter objects it will not be in equilibrium and the absorption rate will be greater than emission rate. As it heats up the emission rate will increase until it matches the absorption rate and equilibrium is reached at a higher temperature.

What about equation 13.19 referenced above?

 

[Dale]

View attachment 243593

You are going to have to be more explicit than that. Please explain the equation

It suggests that net radiation to a blackbody is zero

I am sure it does not suggest that in general.

 

[Nugatory]

What about equation 13.19 referenced above?

You’re misunderstanding something, but without more context I can’t tell you what it is.

 

[UMath1]

In that equation, Ebi is defined as the emission from a perfect blackbody and Ji is the radiosity.

 

[UMath1]

 

[UMath1]

Can someone answer?

 

[Dale]

Not really. I don’t have the book and you seem very secretive about the details. No hint about what the equation is for or what assumptions are made in deriving it. No clear description of the variables. What do you expect? We aren’t mind readers.

If you want help then you have to put a little effort into make your question as complete and clear as possible. Look at what you write through someone else’s eyes.

 

[UMath1]

The equation is used to find the net radiative heat transfer to a given surface. It is typically used when modeling radiation systems as resistive networks. E_bi is the emissivity of the surface were it a blackbody while J_i is the actual radiosity of the surface. ε is the emissivity of the surface. For a blackbody, E_bi = J_i because nothing is reflected and everything is absorbed. Therefore, it would mean that this net radiative heat transfer would yield zero but that doesn't seem to make sense.

 

[Cthugha]

If the net transfer is zero, this does not mean that the system will respond to changes instantaneously. Consider a similar problem that is maybe a bit closer to everyday life.

Consider an empty pipe. At some point you start pouring a bit of water inside continuously. The water will of course flow out of the pipe on the other side. As you increase the amount of water poured inside, the continuity equation demands that also the amount of water that flows out on the other side increases by the same amount. The net amount of water you put into the pipe and that flows out of the pipe will be the same, so there is no net water transfer to the pipe. However, of course this obviously does not mean that the pipe is empty. The reason for this is given by transients. As you increase the amount of water flowing in, the amount of water flowing out does not change instantaneously. For a short amount of time - given by the time it takes for the water to cross the pipe - the amount of water entering the pipe will be larger than the amount of water leaving the pipe. There is not net transfer of water in equilibrium, but only in these transients, when you change the boundary conditions.

Although black bodies react quite quickly to changes in the amount of radiation they receive, their response is not instantaneous. In equilibrium, the net amount of energy coming in and energy going out will be the same, but for the few short instants it takes the system to get to equilibrium if you change the boundary conditions, there will be some energy transfer to or from the system.

 

[Dale]

Ebi is the emissivity of the surface were it a blackbody while Ji is the actual radiosity of the surface.

That isn’t possible. Emissivity is unitless and radiosity is in units of W/m^2. You cannot subtract a radiosity from an emissivity.

Also, what are the subscripts i and b? I assume q is heat in Watts, but it would be nice to not have to assume, and since you haven’t indicated what i means then I really don’t know what qi is.

Also, what are the assumptions made in the derivation of this equation?

 

[Lord Jestocost]

The relevant relations and definitions of the physical quantities can be found in
Electromagnetic, Mechanical, and Transport Properties of Composite ...

"Electromagnetic, Mechanical, and Transport Properties of Composite Materials" by Rajinder Pal

 

[Dale]

For a blackbody, Ebi = Ji because nothing is reflected and everything is absorbed. Therefore, it would mean that this net radiative heat transfer would yield zero but that doesn't seem to make sense.

For a black body both the numerator and the denominator are zero, so the equation is undefined. One of the assumptions in the derivation as given in the link from @Lord Jestocost is that the surface is gray. So it makes sense that the formula would break down for a black body. This is not telling you that a blackbody cannot get hotter, just that you need to use a different equation for a blackbody.

 

[vanhees71]

Summary: If rate of emission = rate of absorption for a blackbody, how can a blackbody undergo temperature change through radiation?

If rate of emission = rate of absorption for a blackbody, how can a blackbody undergo temperature change through radiation? How come black cars get hotter on a sunny day?

The argument goes the other way: If the rate of emission=rate of absorption for a black body, it's in thermal equilibrium with its surrounding and thus its temperature stays constant.

If the sun is shining on your black car, the black body tends to come to thermal equilibrium with the sun (in fact black-body radiation is the mechanism of heat transfer in this case), i.e., it'll heat up. The only reason that fortunately the car doesn't reach the temperature of the sun's photosphere which is several 1000 degrees Kelvin is the fact that it gets cooled by the surrounding air ;-)).