Question

At what rate will energy be emitted from a black body whose filament has a temperature of 3600 K, if at 1800 K the energy is emitted at the rate of 16W?

• A. 32 W
• B. 64 W
• C. 128 W
• D. 256 W

Hint:

The Stefan-Boltzmann Law (\(P \propto T^4\)) shows a very strong dependence on temperature. Doubling the absolute temperature increases the radiated power by a factor of \(2^4 = 16\).

Answer 1

The Correct Option is D

Step 1: State the Stefan-Boltzmann Law. The law states that the total energy radiated per unit surface area of a black body per unit time (the radiant emittance or power, P) is directly proportional to the fourth power of the black body's absolute temperature \(T\). \[ P = \sigma \epsilon A T^4 \] For a given object, \( \sigma, \epsilon, \) and \(A\) are constants, so \( P \propto T^4 \).

Step 2: Set up a ratio for the two different temperatures. Let \(P_1\) be the power emitted at temperature \(T_1\), and \(P_2\) be the power emitted at \(T_2\). \[ \frac{P_2}{P_1} = \frac{T_2^4}{T_1^4} = \left( \frac{T_2}{T_1} \right)^4 \]

Step 3: Substitute the given values and solve for \(P_2\). - \( P_1 = 16 \) W - \( T_1 = 1800 \) K - \( T_2 = 3600 \) K The ratio of temperatures is \( \frac{T_2}{T_1} = \frac{3600}{1800} = 2 \). \[ \frac{P_2}{16} = (2)^4 = 16 \] \[ P_2 = 16 \times 16 = 256 \text{ W} \]