Question

The rate of emission of a perfectly black body at temperature \( 27^\circ C \) is \( E_1 \). If the temperature of the body is raised to \( 627^\circ C \), its rate of emission becomes \( E_2 \). The ratio of \( \frac{E_1}{E_2} \) is:

• A. \( \frac{1}{81} \)
• B. \( \frac{1}{16} \)
• C. \( \frac{1}{25} \)
• D. \( \frac{1}{36} \)
• E. \( \frac{1}{49} \)

Hint:

The Stefan-Boltzmann law states that the rate of emission is proportional to the fourth power of the absolute temperature. Use this relation when comparing emissions at different temperatures.

Answer 1

The Correct Option is A

Step 1: The rate of emission \( E \) of a black body is proportional to the fourth power of its absolute temperature \( T \): \[ E \propto T^4 \] Thus, the ratio of the rates of emission at the two temperatures is: \[ \frac{E_1}{E_2} = \left(\frac{T_1}{T_2}\right)^4 \] Step 2: Convert the temperatures to Kelvin: \[ T_1 = 27^\circ C + 273 = 300 \, {K}, \quad T_2 = 627^\circ C + 273 = 900 \, {K} \] Step 3: Substitute into the equation: \[ \frac{E_1}{E_2} = \left(\frac{300}{900}\right)^4 = \left(\frac{1}{3}\right)^4 = \frac{1}{81} \] Thus, the ratio of \( \frac{E_1}{E_2} \) is \( \frac{1}{81} \).