Answer to Question #245765 in Molecular Physics | Thermodynamics for Cat

Question

A hot tungsten wire filament has a radius of 0.6cm a temperature of 300K and emissivity

0.74. Calculate the rate of energy emission by a 1.0m length of the wire. Ignore the radiation

received from the surrounding environment.

Answer 1

The radiation heat current ($H$) from a body at temperature T (T = 300 K) to its surroundings depends on the mentioned property, the surface area A ($A=2\pi r^2+2\pi rL=2\pi r(r+L)$, that formula is the whole area of the exterior of the wire filament that has a radius r = 0.06 m and length L = 1 m), the emissivity e (e=0.74) and the Stefan-Boltzmann constant ($\sigma = 5.6704\times10^{-8}\frac{W}{m^2 \cdot K^4}$ ).

We proceed to find $H$ after substitution:

$H=Ae\sigma T^4=2\pi r(r+L) e\sigma T^4 \\ H=2\pi [(0.06\,m)(1+0.06)m](0.74)(5.6704\times10^{-8}\frac{W}{m^2 \cdot K^4})(300\,K)^4 \\ H=135.821\,W$

In conclusion, we find that the rate of loss of heat of the hot tungsten wire filament is about 135.821 W.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.