Answer to Question #224808 in Molecular Physics | Thermodynamics for jean

Question

1. Find the steady state temperature of the black body disc having radius 2cm and mass 7g at 45⁰C surrounding temperature, when kept initially at 28⁰C. Hence determine Stefan's constant.

Answer 1

Solution:-

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

Thus, according to the Stefan Boltzmann law

Losses rate

$\frac{dQ_{out}}{dt}=\pi r^2\sigma(T_{sur}^3-T^4)\\\sigma=5.56\times10^{-8}W/m^2K^4\\$

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

Steady state

$\frac{dQ_{out}}{dt}=\frac{dQ_{in}}{dt}$

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

Solution thi equation

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

t=infinite

Put value

$T(t)=T_{sur}$

$$T_{steady}=T_{sur}=273+45=318K=45°C$$