Molar Specific heat of Blackbody radiation

[tanaygupta2000]

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?

 

[Charles Link]

I think you need to compute the blackbody energy density (per unit volume) inside the material.

 

[tanaygupta2000]

I think you need to compute the blackbody energy density (per unit volume) inside the material.

Sir what formula should I use ?

 

[Charles Link]

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.

 

[tanaygupta2000]

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?

 

[Charles Link]

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.

 

[Charles Link]

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.

 

[tanaygupta2000]

Internal energy per unit volume, u = 4σT⁴/c
=> Internal energy, U = u × V_m J
= 4σV_mT⁴/c J
=> Molar specific heat, c_v = ∂U/∂T
= 16σV_mT³/c JK^-1
= 1.02 × 10^-11 JK^-1

 

[Charles Link]

Internal energy per unit volume, u = 4σT⁴/c
=> Internal energy, U = u × V_m J
= 4σV_mT⁴/c J
=> Molar specific heat, c_v = ∂U/∂T
= 16σV_mT³/c JK^-1
= 1.02 × 10^-11 JK^-1

Please check your arithmetic. I get something about 1/20 of what you got.

 

[tanaygupta2000]

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 c_v = 1.02×10^-11/25 = 4.08 × 10^-13 JK^-1

Thank You so much for helping me sir

 

[Charles Link]

Yes sir I made calculation mistake
The final answer is to be divided by 25
Hence c_v = 1.02×10^-11/25 = 4.08 × 10^-13 JK^-1

Thank You so much for helping me sir

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.

 

[tanaygupta2000]

Yes sir I made calculation mistake
The final answer is to be divided by 25
Hence c_v = 1.02×10^-11/25 = 4.08 × 10^-13 JK^-1

Thank You so much for helping me sir

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?

 

[Charles Link]

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.

 

[tanaygupta2000]

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, C_v = 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 ?

 

[Charles Link]

Yes, that is correct, and the blackbody $C_v$ is very much smaller. You can take the ratio to compare them.

 

[Charles Link]

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.