Answer to Question #117044 in Molecular Physics | Thermodynamics for Fahad

Question

calculate steady state temperature of the black body copper disc of radius 4cm and mass 10g. at surrounding temperature of 50 C . when kept initially at 25 C.Also determine stefan boltzmann constant

Answer 1

The disk gains heat $Q_{in}$ due to thermal conductivity from the surrounding and looses heat $Q_{out}$ due to radiation.

Thus, according to the Stefan–Boltzmann law, the looses rate:

$$\dfrac{dQ_{out}}{dt} = \pi r^2\sigma(T_{sur}^4 - T^4),$$

where $T$ is the current temperature of the disk, $\pi r^2$ is the area of the surface of the disk, $\sigma = 5.67\cdot 10^{-8} \dfrac{W}{m^2K^4}$ is the Stefan–Boltzmann constant and $T_{sur}$ is the temperature of the surrounding in Kelvin.

The gain rate will be:

$$\dfrac{dQ_{in}}{dt} = \dfrac{d}{dt}(mc(T - T_{in})) = mc\dfrac{dT}{dt},$$

where $m$ is the mass of the disk, $c$ is the specific heat capacity of the cooper, $T_{in}$ is the initial temperature of the disk in Kelvin.

In the steady state the rate of gain will be equal to the rate of loss, thus:

$$\dfrac{dQ_{out}}{dt} = \dfrac{dQ_{in}}{dt}$$ $$\dfrac{dT}{dt} = \dfrac{\pi r^2\sigma}{mc}T_{sur}^4 - \dfrac{\pi r^2\sigma}{mc}T^4$$

$$\dfrac{dT}{dt} +\dfrac{\pi r^2\sigma}{mc}T^4 = \dfrac{\pi r^2\sigma}{mc}T_{sur}^4$$

The solution of this equation is:

$$T(t) = \dfrac{1}{\sqrt[3]{3\dfrac{\pi r^2\sigma}{mc}t + \dfrac{1}{(T_{start - T_{sur}})^3}}} + T_{sur}$$

In steady state (as $t$ goes to infinity):

$$T_{steady} = T_{sur} = 273+50 = 323K$$

Answer. 323 K or 50 C.