Question

A hot body placed in air 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, find the time in which the body will lose half the maximum heat it can lose.

• A. k/ln(2)
• B. k ln2
• C. $\frac{ln2}{k}$
• D. 2 ln(k)

Answer 1

The correct option is C $\frac{ln2}{k}$

We have,
$\frac{d\theta }{dt}=k(\theta -{\theta }_0)$
Where ${\theta }_0$ is temperature of the surrounding and $\theta$ is the temperature of the body at time t. Suppose $\theta ={\theta }_1$ at time t = 0.
Then,
${}_{{\theta }_1}^{\theta }\frac{d\theta }{\theta -{\theta }_0}=-k_{\theta }^{t}dtor,ln\frac{\theta -{\theta }_0}{{\theta }_1-{\theta }_0}=-ktor,\theta -{\theta }_0=({\theta }_1-{\theta }_0)e^{-kt}\dots \dots \left(1\right)$
This is the maximum heat the body can lose. If the body loses half this heat, the decreases in its temperature will be,
$\frac{\Delta Q_m}{2ms}=\frac{{\theta }_1-{\theta }_0}{2}$
If the body loses this heat in time $t_1$, the temperature at $t_1$ will be,
${\theta }_1-\frac{{\theta }_1-{\theta }_0}{2}=\frac{{\theta }_1+{\theta }_0}{2}$
Putting these value of time and temperature in (1)
$\frac{{\theta }_1+{\theta }_0}{2}-{\theta }_0=({\theta }_1-{\theta }_0)or,e^{-k{t}_1}=\frac{1}{2}or,t_1=\frac{ln2}{k}$