Question

The temperature of an isolated black body falls from $T_1$ to $T_2$ in time $t$. Let $c$ be constant.

• A. $t=c[\frac{1}{T_2}-\frac{1}{T_1}]$
• B. $t=c[\frac{1}{{T_2}^2}-\frac{1}{{T_1}^2}]$
• C. $t=c[\frac{1}{{T_2}^3}-\frac{1}{{T_1}^3}]$
• D. $t=c[\frac{1}{{T_2}^4}-\frac{1}{{T_1}^4}]$

Answer 1

The correct option is C $t=c[\frac{1}{{T_2}^3}-\frac{1}{{T_1}^3}]$

By stefan-boltzmann law, rate of energy radiated by black body is given as: $q=A\sigma \left(T^4\right)\Rightarrow msdT/dt=A\sigma \left(T^4\right)$, where $T$ is the temperature of the body.

So, $dt=C\left(dT/T^4\right)$, where $C$ is the constant of proportionality. Integrating above we get, ${\int }_0^{t}dt=C{\int }_{T_1}^{T_2}dT/T^4$ $\Rightarrow t=(-C/3)\left(1/T^3\right){|}_{T_1}^{T_2}$ $\Rightarrow t=c(\frac{1}{T_2^3}-\frac{1}{T_1^3})$ where $c=-C/3$