# 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?

Homework Helper
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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 ?

Homework Helper
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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.

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tanaygupta2000
Homework Helper
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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.

tanaygupta2000
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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.

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tanaygupta2000
tanaygupta2000
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

Homework Helper
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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

tanaygupta2000
tanaygupta2000
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

Homework Helper
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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
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 cv = 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?

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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, 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 ?