Question

A body cools from $50^{\circ}$ to $45^{\circ}$ in 5 min and to $40^\circ$ in another 8 min. The ambient temperature is:
$\text{A}. \quad 34^{\circ}C$
$\text{B}. \quad 30^{\circ}C$
$\text{C}. \quad 43^{\circ}C$
$\text{D}. \quad 37^{\circ}C$

Answer 1

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:
$\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$ 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 answer:
Here, we can see that the time of consideration is very small hence Newton’s law of cooling is applicable.
For 1 case:
Given, $T_f=45^{\circ}, \ T_i = 50^{\circ} and \ t=5 \ min$.
Hence on putting in the equation, we get:
$\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$
$\dfrac{45-50}{5} = \alpha \left( T - \dfrac{45+50}{2} \right)$
Or $1 = \alpha \left( 47.5 - T \right)$ . . . . ①
Now, again applying for the required case to calculate T :
$\dfrac{T_f-T_i}t = \alpha\left( T_s - T_{avg} \right)$
$\dfrac{40 - 45}{8} = \alpha \left(T - \dfrac{45+40}{2} \right)$
Or $-\dfrac{5}{8} = \alpha \left(T - 42.5 \right)$ . . . .②
Now, dividing equations ① and ②, we get
$\dfrac{1}{-\dfrac58} = \dfrac{47.5-T}{T-42.5}$
Or $8T - 340 = 5T - 237.5$
Hence $T= 34.1^\circ$

So, the correct answer is “Option A”.

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. 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.