Question

An ideal blackbody at temperature \( T \), emits radiation of energy density \( u \). The corresponding value for a material at temperature \( \frac{T}{2} \) is \( \frac{u}{256} \). Its emissivity is ............... (Round off to three decimal places).

Hint:

The emissivity of a material is related to the ratio of the energy density of the material to that of a blackbody at the same temperature.

Answer 1

Correct Answer: 0.06

Step 1: Stefan-Boltzmann law for blackbody radiation.
The energy density of radiation emitted by a blackbody is proportional to the fourth power of its temperature, \[ u = \sigma T^4, \] where \( \sigma \) is the Stefan-Boltzmann constant and \( T \) is the temperature. For a material with emissivity \( \epsilon \), the energy density is given by \[ u_{\text{material}} = \epsilon \sigma T^4. \] Step 2: Relation between temperatures.
For the material at temperature \( \frac{T}{2} \), the energy density becomes \[ u_{\text{material}} = \epsilon \sigma \left( \frac{T}{2} \right)^4 = \epsilon \sigma \frac{T^4}{16}. \] We are given that this value is \( \frac{u}{256} \), so \[ \frac{u}{256} = \epsilon \sigma \frac{T^4}{16}. \] Substitute \( u = \sigma T^4 \) into the equation: \[ \frac{\sigma T^4}{256} = \epsilon \sigma \frac{T^4}{16}. \] Simplify and solve for \( \epsilon \): \[ \frac{1}{256} = \frac{\epsilon}{16} \Rightarrow \epsilon = \frac{16}{256} = 0.0625. \] Final Answer: The emissivity is \( 0.063 \).