Answer to Question #98326 in Astronomy | Astrophysics for Sham Sunder Sainani

Question

A black body is kept at a temperature of 2500K. If volume of body is reduced reversibly and adiabatically to 1/100 of its initial volume what will be the final temperature?

Answer 1

The energy density per unit volume can be calculated according to Wien's radiation law

$u=\frac{8\pi h\nu^3}{c^3}e^{-h\nu/kT},$

and since the volume reduced adiabatically to 1/100 of the initial volume, the energy density increases 100 times, and while other constants ans variables remain the same, the black body acquired the new temperature:

$u_1=100u=\frac{8\pi h\nu_1^3}{c^3}e^{-h\nu_1/kT_1},$

$\frac{u_1}{u}=100=\frac{\nu_1^3}{\nu^3}e^{h(\nu/kT-\nu_1/kT_1)},\\ \space\\ 100\frac{\nu^3}{\nu_1^3}=e^{h(\nu/kT-\nu_1/kT_1)}.$

How to find the frequency of the radiation $\nu$? Wien's displacement law is what will help us:

$\lambda=\frac{c}{\nu}=\frac{b}{T},\\ \space\\ \nu=cT/b,\\ \space\\ 100\frac{T^3}{T_1^3}=e^{h(c/kb-c/kb)}=e^0=1,\\ \space\\ T_1=T\sqrt[3]{100}=2500\sqrt[3]{100}=11604\text{ K}.$

So hot!