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

Question #98326
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?
1
Expert's answer
2019-11-12T17:23:50-0500

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)},\\\\\n\\space\\\\\n100\\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},\\\\\n\\space\\\\\n\\nu=cT\/b,\\\\\n\\space\\\\\n100\\frac{T^3}{T_1^3}=e^{h(c\/kb-c\/kb)}=e^0=1,\\\\\n\\space\\\\\nT_1=T\\sqrt[3]{100}=2500\\sqrt[3]{100}=11604\\text{ K}."

So hot!


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