Answer to Question #248282 in Molecular Physics | Thermodynamics for Sohan

Question

Find the steady-state temperature of the black body disc having radius 3cm and mass 5g at 10 ⁰C surrounding temperature, when kept initially at -10 ⁰C.

Answer 1

Solution;

By Stefan-Boltzmann's law ,the rate of heat loss is given by;

$\frac{dQ_{out}}{dt}=A\sigma(T_s^4-T_d^4)$

Where;

$T_d$ is the temperature of the disc

$T_s$ is the temperature of sorrounding

A is the area of the disc

$\sigma$ is Stefan-Boltzmann's constant

The rate of heat gain will be;

$\frac{dQ_{in}}{dt}=mc\frac{dT}{dt}$

Where m is the mass and c is the specific constant of the disc.

At steady state condition,the rate of heat loss is equal to the rate of heat gain;

Equation both equations;

$mc\frac{dT}{dt}=A\sigma T_s^4-A\sigma T_d^4$

$\frac{dT}{dt}=\frac{A\sigma T_s^4}{mc}-\frac{A\sigma T_d^4}{mc}$

The solution of the equation is;

$T(t)=\frac{1}{\sqrt[3]{\frac{3A\sigma t}{mc}+\frac{1}{(T_d-T_s)^3}}}+T_s$

At steady state;

$t\to\infin$

Hence;

$T_{steady}=T_s$

$T_{steady}=10°c$