Question

The thin tungsten wire with radius r = 0,1 mm and electrical resistivity 8 ρ 25,7 10− = ⋅ Ώ∙m is heated by the current I = 1 А? Which temperature will reach the wire if its heat emission capability equals to a = 0,1 from that of the absolute black body. The Stephan-Boltzmann constant equals 8 σ 5,67 10− = ⋅ Wt/(m2 ⋅К4)

Answer 1

Energy conservation law provides that the power, consumed from the power supply unit is all completely spent on the heat radiation. The Joule-Lentz law provides:

$$P=I^2R=\frac{I^2\rho l}{\pi r^2}$$

The flux of heat radiation equals

$$Ф=М_e2\pi rl$$

The tungsten energy luminosity:

$$M_e=aM_e^0$$

$$\Phi =P=a\sigma T^42\pi rl$$

$$T=\left(\frac{I^2\rho}{2a\sigma \pi^2r^3}\right)^\frac{1}{4}$$

$$T=\left(\frac{1^2(25.7\cdot 10^{-8})}{2(0.1)(5.67\cdot 10^{-8}) \pi^2(0.0001) ^3}\right)^\frac{1}{4}=1200\ K$$