Molar Specific heat of Blackbody radiation

In summary: The conversation discussed how to approach a problem related to a body at temperature T and the dependence of radiative energy per unit area E on the 4th power of T. It was suggested to obtain the expression for specific heat c by differentiating Stefan's law with respect to T. Another approach mentioned was using certain models from Solid State Physics, such as Drude's, Einstein's, or Debye's model. In both cases, only the temperature T was provided and no other constants. It was recommended to compute the blackbody energy density (per unit volume) inside the material using the formula ##\frac{Uc}{4}=\sigma T^4##. To get molar specific energy ##E_m##, this value needs
  • #1
tanaygupta2000
208
14
Homework Statement
A crystalline dielectric solid is at a temperature of 300K. Calculate the contribution of the blackbody radiation to its molar specific heat. Compare it with the classical molar specific heat of 3R.
Relevant Equations
Stefan's law, E = σT^4
Specific heat, c = ∂E/∂T
For a body at temperature T, the radiative energy per unit area E depends on 4th power of T. I can obtain expression for specific heat c by differentiating Stefan's law with respect to T. Would it be the correct way of approaching this problem?
Or do I need to employ certain models from Solid State Physics such as Drude's/Einstein's/Debye's model?
In either cases, I am provided with only the temperature T and no other constants. How should I attempt such a problem?
 
Physics news on Phys.org
  • #2
I think you need to compute the blackbody energy density (per unit volume) inside the material.
 
  • #3
Charles Link said:
I think you need to compute the blackbody energy density (per unit volume) inside the material.
Sir what formula should I use ?
 
  • #4
See https://www.physicsforums.com/threa...n-plancks-law-derivation.961396/#post-6097855
To simplify things for you, blackbody energy density (per unit volume) ## U ## satisfies ##\frac{Uc}{4}=\sigma T^4 ##. I need to double check and/or verify this formula, but I think I have it correct. Yes= See http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/raddens.html
To get molar specific energy ##E_m ##, you need to multiply ## U ## by the molar volume ##V_m ##. Finally, to get molar specific heat, you take the derivative of the molar specific energy w.r.t. to T.
 
Last edited:
  • #5
Charles Link said:
See https://www.physicsforums.com/threa...n-plancks-law-derivation.961396/#post-6097855
To simplify things for you, blackbody energy density (per unit volume) ## U ## satisfies ##\frac{Uc}{4}=\sigma T^4 ##. I need to double check and/or verify this formula, but I think I have it correct. Yes= See http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/raddens.html
Upon substituting, I am getting Uc = 4σT4 = 4 × (5.67×10-8) × (300)4 = 1837.08
But there is no relation with R.
Is this correct?
 
  • #6
See the additions to post 4. Meanwhile you need to compute ## U ##. The speed of light ## c=3.00 E+8 ##, and take the derivative w.r.t. T before plugging in T=300 K.
 
  • Like
Likes tanaygupta2000
  • #7
For some rough estimates, you can estimate an atomic size as (2E-10)^3 m^3, so that a molar volume will be ##V_m = (6.02E+23 )(8E-30) =## 5E-6 m^3 which is 5 cm^3=probably a good estimate.
Meanwhile one mole of water is 18 grams and occupies 18 cm^3, so we are in the right ballpark with the number 5 E-6 m^3. For another example, aluminum has density of 2.7 g/cm^3 with atomic weight of 27, so one mole will have a volume of 10 cm^3.
 
Last edited:
  • Informative
Likes tanaygupta2000
  • #8
Internal energy per unit volume, u = 4σT4/c
=> Internal energy, U = u × Vm J
= 4σVmT4/c J
=> Molar specific heat, cv = ∂U/∂T
= 16σVmT3/c JK-1
= 1.02 × 10-11 JK-1
 
  • #9
tanaygupta2000 said:
Internal energy per unit volume, u = 4σT4/c
=> Internal energy, U = u × Vm J
= 4σVmT4/c J
=> Molar specific heat, cv = ∂U/∂T
= 16σVmT3/c JK-1
= 1.02 × 10-11 JK-1
Please check your arithmetic. I get something about 1/20 of what you got.
 
  • Like
Likes tanaygupta2000
  • #10
Charles Link said:
Please check your arithmetic. I get something about 1/20 of what you got.
Yes sir I made calculation mistake
The final answer is to be divided by 25
Hence cv = 1.02×10-11/25 = 4.08 × 10-13 JK-1

Thank You so much for helping me sir :smile:
 
  • Like
Likes Charles Link
  • #11
tanaygupta2000 said:
Yes sir I made calculation mistake
The final answer is to be divided by 25
Hence cv = 1.02×10-11/25 = 4.08 × 10-13 JK-1

Thank You so much for helping me sir :smile:
Meanwhile, what do you get for ## C_v=3 R ##, (the classical specific heat)?
The blackbody part is clearly negligible for any practical purposes, as the calculation will show.
 
  • #12
tanaygupta2000 said:
Yes sir I made calculation mistake
The final answer is to be divided by 25
Hence cv = 1.02×10-11/25 = 4.08 × 10-13 JK-1

Thank You so much for helping me sir :smile:
Sir there is one last doubt
We have calculated the molar volume, so it will have units m3 mol-1
so the final answer is 4.08 × 10-13 JK-1 mol-1

Can I write it as 0.49R?
 
  • #13
The answer ##C_v= 4 E-13 ## we have is for the blackbody part of the "molar" specific heat. Often they place a bar over ##C_v ## or a subscript ##m ## if it is for just one mole. Otherwise, yes, the units on ##C_V ## are joules/(mole-Kelvin). ## \\ ##
Meanwhile the classical molar specific heat is ##C_v=3R ##. The ##R ## is the universal gas constant. You may know it as ##R=.08206 ## liters-atmospheres/(mole-Kelvin), or ##R=1.987 ## calories/(mole-Kelvin). With 1 calorie=4.184 joules, you can compute what ##C_v=3R ## is.
 
  • #14
Charles Link said:
The answer ##C_v= 4 E-13 ## we have is for the blackbody part of the "molar" specific heat. Often they place a bar over ##C_v ## or a subscript ##m ## if it is for just one mole. Otherwise, yes, the units on ##C_V ## are joules/(mole-Kelvin). ## \\ ##
Meanwhile the classical molar specific heat is ##C_v=3R ##. The ##R ## is the universal gas constant. You may know it as ##R=.08206 ## liters-atmospheres/(mole-Kelvin), or ##R=1.987 ## calories/(mole-Kelvin). With 1 calorie=4.184 joules, you can compute what ##C_v=3R ## is.
Sir the classical value is, Cv = 3×8.3136 JK-1mol-1
= 24.94 JK-1mol-1
Is this right ?
Is the question asking to compare this value with the value which we have calculated ?
 
  • Like
Likes Charles Link
  • #15
Yes, that is correct, and the blackbody ##C_v ## is very much smaller. You can take the ratio to compare them.
 
  • Like
Likes tanaygupta2000
  • #16
What this is saying is virtually all of the heat capacity of a solid at room temperature is do to vibrations of the atoms in the lattice, and not affected by the very small amount of energy of the blackbody type electromagnetic radiation inside the solid. (as the temperature increases, there is more vibrational energy. To raise the temperature you need to add energy because of the increased vibrations).There is also a small component due to motion of the electrons, and in a more advanced course they will teach you that this component is also rather small. ## \\ ##One additional item worth mentioning is that the blackbody radiation from the surface of a solid comes from the vibrations of the atoms, and unless you keep putting additional energy into the solid, the vibrational energy will decrease as the solid radiates, and the solid will become cooler. To try to clarify this, the vibrations are referred to as "phonon" modes. When we computed the blackbody energy density ## U ## above, we computed the energy of the "photon" modes. We showed that the phonon modes contain much more energy than the photon modes.
 
Last edited:
  • Informative
Likes tanaygupta2000

Related to Molar Specific heat of Blackbody radiation

1. What is the molar specific heat of blackbody radiation?

The molar specific heat of blackbody radiation is the amount of heat required to raise the temperature of one mole of a blackbody substance by one degree Celsius.

2. How is the molar specific heat of blackbody radiation measured?

The molar specific heat of blackbody radiation is typically measured using calorimetry, where the change in temperature of a known amount of the substance is measured after a certain amount of heat is applied.

3. What is the significance of the molar specific heat of blackbody radiation?

The molar specific heat of blackbody radiation is an important property in thermodynamics and is used to calculate the amount of heat required for a substance to undergo a change in temperature. It also provides insights into the energy distribution of a blackbody substance.

4. How does the molar specific heat of blackbody radiation vary with temperature?

The molar specific heat of blackbody radiation is temperature-dependent and typically increases with increasing temperature. This is due to the increase in the number of vibrational modes of the substance at higher temperatures.

5. Can the molar specific heat of blackbody radiation be negative?

No, the molar specific heat of blackbody radiation cannot be negative. It is a positive value that represents the amount of heat required to raise the temperature of the substance. A negative value would imply that the substance releases heat when its temperature increases.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
194
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
792
  • Introductory Physics Homework Help
Replies
3
Views
1K
Replies
29
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
0
Views
215
Back
Top