A gas of photons in equilibrium is referred to as black body radiation. One can

Question

A gas of photons in equilibrium is referred to as black body radiation. One can show that at temperature T, a gas of photons in a box of volume V has energy

E= sigma *V *T^4

Here (sigma=7.56*10^-15) is the Stefan-Boltzman constant.

(a) If the temperature of the gas changes by deltaT while the volume remains fixed, what is the heat flow into the box?

(b) Why is the change in the entropy deltaS, assuming that the change is done slowly?

(c) Inegrate the relation found in (b) to find S as a function of V, T. (for the limits of integration, you can take the lower limit to be T=0, and just call the upper limit T.)

(d) What is S as a function of V, E?

(e) Finally take the expression S (E,V) found in (d) as your starting point. Use the definition of temperature to compute T from S(E,V). This should bring you back to the beginning of the problem.

Explanation / Answer

a)
dE =4 sigma *V *T^3*dT

is the energy(heat) flow

b) d(dS) = d(dE/T)

integrating both sides

dS = dE/T =
4sigma *V *T^2*dT

c) integrating above expression

S = 4(sigma *V *T^3)/3

d) S = 4E/3T , T = (E/Vsigma)^1/4 >>> S = 4(E^3/4)/3((V*sigma)^1/4)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.