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1. |
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» (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`. |
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