Question

The emissivity of tungsten is approximately 0.35 A tungsten sphere 1 cm in radius is suspended within a large evacuated enclosure whose walls are at 300 K. What power input is required to maintain the sphere at a temperature of 3000 K? ( \[\sigma = 5.67 \times 10 - 8\] inSI unit)
A) 1020 W
B) 2020 W
C) 3020 W
D) 4020 W

Answer 1

Hint: Amount of heat radiated by tungsten will be equal to the amount of energy supplied. According to Stephan’s law the heat loss is given by, \[e\sigma A({T^2} - {T_0}^2)\] . Substitute the values of initial and final temperature, area, emissivity and the constant. Simplify to find the value power required.

Complete step-by-step solution
Energy is never produced or destroyed in this universe; there is always a transfer of energy from one form to another. Energy loss can be in the form of heat. In our case also, the amount of heat to be supplied per unit time to maintain the temperature of the sphere is the heat lost per unit time by that sphere.
Using Stephan’s law,
 \[
  Heat\,lost\,of\,radiation = e\sigma A({T^2} - {T_0}^2) \\
  T = 3000k \\
  {T_0} = 300k \\
  A = 4\pi {r^2} = 4\pi \times {0.01^2} \\
  Substituting, \\
  P = 0.35 \times 5.67 \times {10^{ - 8}} \times 4\pi {(0.01)^2}[{3000^2} - {300^2}] \\
  P = 2019.8W \\
 \]

Therefore, the correct answer is option B.

Note: This equation is in terms of power or heat transferred per unit time. If you want to calculate the heat transferred in say 1 minute, then multiply the power value with 60s.According to Stephan's law the radiant energy emitted by a perfectly black body per unit area per second is directly proportional to the fourth power of its absolute temperature.
 $E \propto {T^4} \Rightarrow E = \sigma {T^4}$