Understanding Thermal Radiation: Entropy, Gibbs Function, and Heat Capacity

In summary, the conversation is discussing the thermodynamic treatment of thermal radiation as a gas of photons with internal energy and pressure. The goal is to show that entropy density, Gibbs function, and heat capacity at constant volume and pressure can be calculated using the energy density. The conversation includes a request for help with finding Cv and a suggestion to use a specific equation to find the values in terms of T.
  • #1
bon
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Homework Statement



Thermal radiation can be treated thermodynamically as a gas of photons with internal energy U = u(T) V and pressure p = u(T)/3 where u(T) is the energy density. Show that

(a) entropy density s is given by s = 4p/T
(b) Gibbs function = 0
(c) heat capacity at constant volume Cv = 3s per unit volume
(d) heat capacity at constant pressure Cp is infinite.


Homework Equations





The Attempt at a Solution



I've done (a), (b), but am having trouble on (c). I guess I am trying to find T(Ds/DT) where capital D is partial.. but I am having touble arriving at 3s..

any hints? Thanks! :)
 
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  • #2
Can I see your work up to this point for c)? I can't figure out where your going wrong unless I can see your calculation.
 
  • #3
G01 said:
Can I see your work up to this point for c)? I can't figure out where your going wrong unless I can see your calculation.

Ok sure so for (c) I need Cv

Cv = T (DS/DT)v (i.e. at constant volume)

I guess I am trying to work out Cv/V i.e. heat capacity per unit volume

Now since V is constant and S = sV

I can write Cv = T(Ds/DT)v

I just can't seem to see where to go from here..whatever i try i can't get 3s out... :S
 
  • #4
any ideas?
 
  • #5
OK. Here's something to get you started:

I don't know the constants off the top of my head, but let's just call them all A:

[tex]u(T)=AT^4[/tex]

Using this, what is p in terms of only T? Then, what is s in terms of only T?

Take the derivative of this result and plug it into the equation for Cv. You should then be able to do some algebra to show it is equal to 3 times the result you got for s in terms of T.
 
  • #6
Thanks but how can i get to this using only the results I've been given?
 
  • #7
a-n-y-o-n-e?
 

FAQ: Understanding Thermal Radiation: Entropy, Gibbs Function, and Heat Capacity

What is Quick Thermal Physics q?

Quick Thermal Physics q is a measurement that represents the rate of thermal energy transfer, or heat flow, in a given system. It is calculated by dividing the amount of thermal energy transferred by the time it takes for the transfer to occur.

How is Quick Thermal Physics q related to temperature?

Quick Thermal Physics q is directly proportional to the temperature difference between two systems. This means that as the temperature difference increases, so does the rate of thermal energy transfer.

What are some real-life examples of Quick Thermal Physics q?

Some examples of Quick Thermal Physics q in action include the transfer of heat from a hot stove to a pot of water, the cooling of a hot cup of coffee, and the melting of an ice cube in a warm room.

How can Quick Thermal Physics q be measured?

There are various instruments that can be used to measure Quick Thermal Physics q, such as calorimeters, thermometers, and infrared cameras. These instruments allow scientists to accurately measure the amount of thermal energy transferred in a given system.

What factors affect Quick Thermal Physics q?

The rate of thermal energy transfer, or Quick Thermal Physics q, can be affected by several factors including the temperature difference between two systems, the material properties of the objects involved, and the presence of any insulating materials.

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