Question

If the emission rate of blackbody at $0^\circ\text{C}$ is $R$ then, the rate of emission at $273^\circ\text{C}$ is

• A. \(2R \)
• B. \(4R \)
• C. \(8R \)
• D. \(16R \)
• E. \(32R \)

Hint:

In radiation problems, temperature must always be in Kelvin. Doubling the Celsius temperature does not mean doubling the absolute temperature.

Answer 1

The Correct Option is D

Concept: The Stefan-Boltzmann Law states that the power radiated per unit area of a blackbody (emission rate) is proportional to the fourth power of its absolute temperature: \[ R = \sigma T^4 \] where:
• \(R\) is the rate of emission.
• \(\sigma\) is the Stefan-Boltzmann constant.
• \(T\) is the temperature in Kelvin.

Step 1: Convert temperatures to Kelvin.
To use the radiation law, Celsius must be converted to Kelvin using \(T(K) = T(^\circ\text{C}) + 273\). Initial temperature \(T_1 = 0^\circ\text{C} + 273 = 273 \, \text{K}\). Final temperature \(T_2 = 273^\circ\text{C} + 273 = 546 \, \text{K}\).

Step 2: Determine the ratio of emission rates.
Since \(\sigma\) is constant, the ratio of the rates is: \[ \frac{R_2}{R_1} = \left( \frac{T_2}{T_1} \right)^4 \]

Step 3: Calculate the final rate \(R_2\).
Substitute the values: \[ \frac{R_2}{R} = \left( \frac{546}{273} \right)^4 = (2)^4 \] \[ R_2 = 16R \]