Question

The temperature of an isolated black body falls from $T_1$ to $T_2$ in time $t$, then $t$ is (Let $c$ be a constant)

• A. $t=c(\frac{1}{T_2}-\frac{1}{T_2})$
• 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

Answer: (C)

$t=c(\frac{1}{T_2^3}-\frac{1}{T_1^3})$

According to Newton's law of cooling

The rate of heat loss is given by

$\frac{dQ}{dt}=ms\frac{dT}{dt}$

where $m$ and $s$ are mass and specific heat of the body, $\frac{dT}{dt}$ is rate of change in temperature.

As the black body is isolated the total heat generated or loss by the black body will radiate as fourth power of temperature of the body.

That is

$\frac{dQ}{dt}=-\sigma \epsilon A{T}^4$

By equating the above two equations

$ms\frac{dT}{dt}=-\sigma \epsilon A{T}^4$

$\frac{dT}{T^4}=-\frac{\sigma \epsilon A}{ms}dt$

given that the temperature of the body falls from $T_1$ to $T_2$ in time $t$

by integrating the above equation on both sides and applying the limits for temperature and time, we get the time taken for the body to decrease its temperature from $T_1$ to $T_2$

${\int }_{T_1}^{T_2}\frac{dT}{T^4}=-\frac{\sigma \epsilon A}{ms}{\int }_0^{t}dt-\left(\frac{1}{3}\right)(\frac{1}{T_2^3}-\frac{1}{T_1^3})=-\frac{\sigma \epsilon A}{ms}tt=\frac{ms}{3\sigma \epsilon A}(\frac{1}{T_2^3}-\frac{1}{T_1^3})=c(\frac{1}{T_2^3}-\frac{1}{T_1^3})$

option (C) is the correct answer