(d) Calculate the maximum amount of heat which may be lost per second by radiation from a sphere of 5 cm in diameter at a temperature of 600 K when is placed in an encloser at a temperature of 300 K. Given that \(\sigma = 5.7 \times 10^{-12} \, \text{watts/cm}^{-2} / (\circ C)^{-4}\).

Another Explanation (5): Here's the solution: The formula for radiative heat transfer is given by Stefan-Boltzmann law: \(P = A \sigma \epsilon (T_1^4 - T_2^4)\) where: * \(P\) is the power radiated (heat lost per second) in watts * \(A\) is the surface area of the sphere in \(cm^2\) * \(\sigma\) is the Stefan-Boltzmann constant (\(5.7 \times 10^{-12} \, \text{watts/cm}^{-2} K^{-4}\)) * \(\epsilon\) is the emissivity (assumed to be 1 for a perfect blackbody) * \(T_1\) is the temperature of the sphere in Kelvin (600 K) * \(T_2\) is the temperature of the enclosure in Kelvin (300 K) First, calculate the surface area of the sphere: \(A = 4\pi r^2 = 4\pi (2.5 \, cm)^2 \approx 78.54 \, cm^2\) Now, substitute the values into the Stefan-Boltzmann law: \(P = 78.54 \, cm^2 \times 5.7 \times 10^{-12} \, \frac{watts}{cm^2 K^4} \times 1 \times (600^4 - 300^4) \, K^4\) \(P \approx 78.54 \times 5.7 \times 10^{-12} \times (1296 \times 10^8 - 81 \times 10^8) \, watts\) \(P \approx 78.54 \times 5.7 \times 10^{-12} \times 1215 \times 10^8 \, watts\) \(P \approx 0.547 \, watts\) Therefore, the maximum amount of heat lost per second by radiation is approximately 0.547 watts.