Question

A cavity of volume `V = 1.01` is filled with thernal radiation at a temperature `T = 1000K`. Find:
(a) the heat capacity `C_(v)`, (b) the entropy `S` of that radiation.

Answer 1

(a) The total internal energy of the cavity is
`U = (4 sigma)/(c )T^(4)V`
Hence `C_(v) = ((delU)/(delT))_(v) = (16 sigma)/(C )T^(3)V`
`= (16xx5.67xx 10^(8))/(3 xx 10^(8)) xx 10^(9) xx 10^(-3)"Joule"//^(@)K`
`= (1.6 xx 5.67)/(3)nJ//K = 3.024nJ//K`
(b) From first law
`TdS = dU + pdV`
`= VdU + UdV + (u)/(3)dV (p = (U)/(3))`
`= VdU+(4U)/(3)dV`
`= (16sigma)/(C)VT^(3)dT +(16sigma)/(3C)T^(4)dV`
or `dS = (16sigma)/(C)VT^(2)dT +(16sigma)/(3C)T^(3)dV`
`= ((16sigma)/(3C)VT^(3))`
Hence `S = (16sigma)/(3C)VT^(3) = (1)/(3)C_(v) = 1.008nJ//K`.