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There are two cavities (Fig.) with small holes of equal diameters d = 1.0 cm and perfectly reflecting outer surfaces. The distance between the holes is l = 10 cm. A constant temperature T_(1) = 1700K is maintained in cavity 1. Calculate the steady-state temperature inside cavity 2. |
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Answer» Solution :Taking account of cosine low of emission we write for the enegry radiated per second by the hole in cavity `# 1` as `di(OMEGA) = A cos thetad Omega` where `A` is an consatnt, `d Omega` is an element of SOLID angle around some DIRECTION defined by the symbol `Omega`. Integrating over the whole forwed hemisphere we get `I = A underset(0)overset(pi//2)int cos theta2pi sin theta d theta = pi 4`  We find `A` by equating this to the quantity `sigmaT_(1)^(4).(pid^(2))/(4)sigma` is stefan-Boltzman constant and `d` is the diameter of the hole. Then `A = (1)/(4) sigma d^(2)T_(1)^(4)` Now energy reaching `2` form `1` is `(cos theta = 1)` `(1)/(4)sigmad^(2)T_(1)^(4).Delta Omega` where `Delta Omega = ((pid^(2)//4))/(l^(2))`is the solid angle substended by the hole of `2` at `1`. {We are ASSUMING `dlt lt l` so `Delta Omega =` area of hole`//("disatnce")^(2)}`. This must equal `sigmaT_(2)^(4)pid^(2)//4` which is the energy EMITTED by `2`. Thus equating `(1)/(4)sigmad^(2)T_(1)^(4) (pid^(2))/(4l^(2)) = sigmaT_(2)^(4) (pid^(2))/(4)` or `T_(2) = T_(1) sqrt((d)/(2l))` Substituting we get `T_(2) = 0.380kK = 380 K`. |
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