Question

A hot body placed in air is cooled down according to Newton’s law of cooling, the rate of decrease of temperature being $k$ times the temperature difference from the surrounding. Starting from $t=0$, the time in which the body will lose half the maximum heat it can lose is $\frac{\ln a}{k}$. Find the value of $a$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Let $T_0$ be the temperature of the surrounding and $T$ be the temperature of the body at time $t$.

According to the given question,

$\frac{dT}{dt}=-k(T-T_0)$

Let the initial temperature of body be $T_1$

$\Rightarrow {\int }_{T_1}^T\frac{dT}{T-T_0}={\int }_0^t-kdt$

$\Rightarrow {\left[\ln \right(T-T_0\left)\right]}_{T_1}^T=[-kt{]}_0^t$

$\Rightarrow ln\left[\frac{T-T_0}{T_1-T_0}\right]=-kt$

$\Rightarrow \frac{T-T_0}{T_1-T_0}=e^{-kt}$

$\Rightarrow T=T_0+(T_1-T_0)e^{-kt}........\left(1\right)$

As we know that the body continues to lose heat till its temperature becomes equal to that of the surrounding, so the maximum loss of heat will be

$Q_{max}=ms(T_1-T_0)$

Where, $s$ is specific heat capacity of the body.

Let $T^{\prime }$ be the temperature at which the body loses half of $Q_{max}$.

$\Rightarrow \frac{Q_{max}}{2}=ms(T_1-T^{\prime })$

$\Rightarrow \frac{ms(T_1-T_0)}{2}=ms(T_1-T^{\prime })$

$\Rightarrow T^{\prime }=T_1-\left(\frac{T_1-T_0}{2}\right)=\frac{T_1+T_0}{2}$

Substituting $(T=T^{\prime })$ in equation $\left(1\right)$,

$\frac{T_1+T_0}{2}=T_0+(T_1-T_0)e^{-kt}$

$\Rightarrow \frac{T_1-T_0}{2}=(T_1-T_0)e^{-kt}$

$\Rightarrow e^{-kt}=\frac{1}{2}$

$\Rightarrow kt=\ln 2$

$\Rightarrow t=\frac{\ln 2}{k}$

$\therefore$ Value of $a$ is $2$.

Hence, option $\left(\text{B}\right)$ is correct.