Consider black body radiation in a cubical bo | vedclass

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(A) The energy density $u$ of black body radiation in a cavity at temperature $T$ is given by $u = aT^4$,where $a$ is the radiation constant.The total

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halves

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(A) The energy density $u$ of black body radiation in a cavity at temperature $T$ is given by $u = aT^4$,where $a$ is the radiation constant.The total energy $E$ contained in the box is the product of energy density $u$ and the volume $V$ of the box.$E = u 	imes V = aT^4 	imes V$.Initially,the side length is $L$,so the volume $V_1 = L^3$. The energy is $E_1 = aT^4 L^3$.Finally,the side length is doubled,so $L' = 2L$,and the new volume $V_2 = (2L)^3 = 8L^3$.The temperature is halved,so $T' = T/2$.The new total energy $E_2 = a(T')^4 V_2 = a(T/2)^4 (8L^3)$.$E_2 = a(T^4 / 16) (8L^3) = (8/16) aT^4 L^3 = (1/2) aT^4 L^3$.$E_2 = E_1 / 2$.Therefore,the total energy halves.

Consider black body radiation in a cubical box at absolute temperature $T$. If the length of each side of the box is doubled and the temperature of the walls of the box and that of the radiation is halved,then the total energy

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