Question

A cup of tea cools from $80^{\circ}$to$60^{\circ}$ in 1 min. The ambient temperature is$30^{\circ}$. In next 1 minute, the temperature will be:
$\text{A}. \quad 40^{\circ}C$
$\text{B}. \quad 45^{\circ}C$
$\text{C}. \quad 48^{\circ}C$
$\text{D}. \quad 42^{\circ}C$

Answer 1

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.

Formulas used:
$\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$ 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.

Complete step by step answer:
Here, we can see that the time of consideration is very small and also the temperature of surrounding is comparable with the temperature of the body, hence Newton’s law of cooling is applicable.
Given, $T_f=60^{\circ}, \ T_i = 80^{\circ} , t=1 \ min \ and \ T_s = 30^{\circ}$.
Hence on putting in the equation, we get:
$\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$
$\dfrac{60-80}{1} = \alpha \left( 30 - \dfrac{60+80}{2} \right)$
Or $\alpha = \dfrac{-20}{-40}=\dfrac12$
Hence we got the value of constant in the equation.
Now, again applying for the required case to calculate $T_f$:
$\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$
$\dfrac{T - 60}{1} = \dfrac12 \left(30 - \dfrac{T+60}{2} \right)$
[ One can also apply the equation w.r.t $80^{\circ}$ by taking time as 2 min and the result will be nearly same ]
Or $2T - 120 = 30 - \dfrac T2 - 30 = - \dfrac T2$
Or, $\dfrac {5T}{2} = 120$
Or, $T = 48^{\circ}C$
Hence option C is correct.

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-Boltzmaan’s law.
Also $\alpha$ is a constant in this equation. But actually it depends upon certain factors. Infact $\alpha =\dfrac{ 4\sigma eAT_{\circ}^3\Delta T}{ms}$ [where symbols have their usual meaning] depends upon the surrounding temperature also. So alpha is constant only till the medium is not changed.