Question

Calculate the energy radiated in one minute by a perfectly black body of surface area 200 cm\(^2\) when it is maintained at 127\(^{\circ}\)C.

Hint:

Always convert units before calculating! Area must be in \(m^2\), time in seconds, and most importantly, temperature must be in Kelvin (\(K = ^{\circ}C + 273\)) for the Stefan-Boltzmann law.

Answer 1

According to the Stefan-Boltzmann law, the power (P) or rate of energy radiated by a perfectly black body is given by: \[ P = \sigma A T^4 \] where \(E = P \times t\) is the total energy radiated in time \(t\). So, \(E = \sigma A T^4 t\).
Given:
Stefan's constant, \(\sigma \approx 5.67 \times 10^{-8} \, W/m^2K^4\)
Surface area, \(A = 200 \, cm^2 = 200 \times 10^{-4} \, m^2 = 2 \times 10^{-2} \, m^2\)
Temperature, \(T = 127^{\circ}C = 127 + 273 = 400 \, K\)
Time, \(t = 1 \, minute = 60 \, s\)
Now, calculate the energy radiated (E): \[ E = (5.67 \times 10^{-8}) \times (2 \times 10^{-2}) \times (400)^4 \times 60 \] \[ E = (5.67 \times 10^{-8}) \times (2 \times 10^{-2}) \times (256 \times 10^8) \times 60 \] \[ E = 5.67 \times 2 \times 256 \times 10^{-2} \times 60 \] \[ E = 11.34 \times 256 \times 0.6 \] \[ E \approx 2903 \times 0.6 \] \[ E \approx 1741.8 \, J \] The energy radiated in one minute is approximately 1742 Joules.