Question

Two black bodies P and Q have equal surface areas and are kept at temperatures \( 127^\circ C \) and \( 27^\circ C \) respectively. The ratio of thermal power radiated by A to that by B is:

• A. 81 : 256
• B. 177 : 127
• C. 127 : 177
• D. 256 : 81

Hint:

When comparing the power radiated by two bodies, remember that the power is proportional to the fourth power of the temperature. Convert the temperatures to Kelvin before applying the Stefan-Boltzmann law.

Answer 1

The Correct Option is D

The power radiated by a black body is given by Stefan-Boltzmann law: \[ P = \sigma A T^4 \] where: - \( P \) is the power radiated, - \( \sigma \) is the Stefan-Boltzmann constant, - \( A \) is the surface area, and - \( T \) is the absolute temperature in Kelvin. Since the surface areas of bodies P and Q are the same, the ratio of the power radiated by the bodies depends only on their temperatures: \[ \frac{P_{\text{P}}}{P_{\text{Q}}} = \frac{T_{\text{P}}^4}{T_{\text{Q}}^4} \] Now, converting the temperatures to Kelvin: \[ T_{\text{P}} = 127^\circ C + 273 = 400 \, \text{K} \] \[ T_{\text{Q}} = 27^\circ C + 273 = 300 \, \text{K} \] Now substitute these values into the ratio: \[ \frac{P_{\text{P}}}{P_{\text{Q}}} = \frac{400^4}{300^4} = \left( \frac{400}{300} \right)^4 = \left( \frac{4}{3} \right)^4 = \frac{256}{81} \] Thus, the ratio of the power radiated by P to that by Q is \( 256 : 81 \).