(d) An aluminum foil of relative emittance 0.1 is placed in between two concentric spheres at temperatures 300 K and 200 K respectively. Calculate the temperature of the foil after the steady state is reached. Assume that the spheres are perfect blackbody radiators. Also calculate the rate of energy transfer between one of the spheres and the foil. [\(\sigma = 5.672 \times 10^{-8} \, \text{M.K.S. units}\)]

Another Explanation (5): Let \(T_1 = 300 K\) and \(T_2 = 200 K\) be the temperatures of the outer and inner spheres respectively. The relative emittance of the aluminum foil is ε = 0.1. At steady state, the heat transfer rate from the outer sphere to the foil equals the heat transfer rate from the foil to the inner sphere. The heat transfer rate between two blackbodies is given by the Stefan-Boltzmann law: \(Q = A \sigma (T^4 - T'^4)\), where A is the area, σ is the Stefan-Boltzmann constant, and T and T' are the temperatures. Let T_f be the temperature of the foil. Assuming the foil radiates to both spheres, the energy balance is: \(A \sigma (T_1^4 - T_f^4) \epsilon = A \sigma (T_f^4 - T_2^4) \epsilon\) Since the area is the same for both sides of the foil, we can simplify: \(T_1^4 - T_f^4 = T_f^4 - T_2^4\) \(2T_f^4 = T_1^4 + T_2^4\) \(T_f^4 = \frac{T_1^4 + T_2^4}{2}\) \(T_f = \sqrt[4]{\frac{300^4 + 200^4}{2}} \approx 259.6 K\) The rate of energy transfer between the outer sphere and the foil is: \(Q = A \epsilon \sigma (T_1^4 - T_f^4) = A (0.1)(5.672 \times 10^{-8}) (300^4 - 259.6^4) \approx 573.5 A \, W\) Where A is the surface area of the foil. The same amount of energy is transferred between the foil and the inner sphere.