Hot coffee in a mug cools from $90^{\circ}$C to $70^{\circ}$C in 4.8 minutes. The room temperature is $20^{\circ}$C. Applying newton’s law of cooling the time needed to cool it further by $10^{\circ}$C should be nearly:
A. 4.2 min
B. 3.8 min
C. 3.2 min
D. 2.4 min

Hint: Newton was the first successful person to compute the relation between temperature variation and time. However his thesis were limited only for some special type of system in which certain assumptions were made. The assumptions made so that the law is valid are that the temperature difference of the body must not be very large and the time of consideration of the variation also must be small. This law is called Newton’s law of cooling.
Formula used:
$\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$

Complete answer:
Since we are given here that the law is applicable, hence we can use Newton's law of cooling. Hence using: $\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$ where $T_f, T_i \ and \ T_s$are the final, initial and surrounding temperature with respect to the body. Also, $T_{avg} = \dfrac{(T_f+T_i)}2 \ and \ \alpha$is a constant.
Given (Case I): $T_f = 70^{\circ},\ T_i = 90^{\circ},\ t = 4.8 \ min \ and \ T_s = 20^{\circ}$
Thus;
$\dfrac{70-90}{4.8} = \alpha\left( 20 - \dfrac{90+70}{2} \right)$
$\implies \alpha = \dfrac{20}{4.8\times 60} = \dfrac{1}{14.4}$
Now, for the second case:
$\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$
$\dfrac{60-70}{t} = \dfrac{1}{14.4}\left( 20 - \dfrac{70+60}{2}\right)$
$\implies t = \dfrac{-144}{-45} = 3.2\ min$

So, the correct answer is “Option C”.

Note:
It should be noted that this form of Newton’s law of cooling is the most approximate form of the law. Important point to be remembered using it is that the time and temperature difference must not vary by a certain amount i.e. should be small. If we want to calculate the temperature over a longer period of time and to any value of temperature, one can use Stephen-Boltzmann’s law.

Also $\alpha$ is a constant in this equation. But actually it depends upon certain factors. In fact $\alpha =\dfrac{ 4\sigma eAT_{\circ}^3}{ms}$ [where symbols have its usual meaning ]depends upon the surrounding temperature also. So alpha is constant only till the medium is not changed.