Question

A body initially at 800 C cools to 640C in 5 minutes and to 520 in 10 minutes. The temperature of the surrounding is ____________
A) 260C
B) 160C
C) 360C
D) 400C

Answer 1

We can solve this question using Newton’s law of cooling. It is applied to both warming and cooling. Newton’s law of cooling is helpful to understand the rate of cooling. i.e. how fast the hot object is cooling.

Complete step by step answer:
Let us divide this question in two parts. (i) from 800C to 640C and (ii) from 640C to 520C. Now, let us analyse the question in detail.
As per Newton’s law of cooling the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings.
$\dfrac{\theta_1-\theta_2}{t}=K[\dfrac{\theta_1+\theta_2}{2}-\theta_0]$
Where $\theta_1$ initial temperature and $\theta_2$ is the final temperature, K is the constant and $\theta_0$ is the room temperature.
Let us solve step by step,
Consider first case, $\dfrac{80-64}{5}=K[\dfrac{80+64}{2}-\theta_0]$
$\dfrac{16}{5}=K[\dfrac{144}{2}-\theta_0] 3.2=K[72-\theta_0]$ --------(1)
Now, let us discuss the second case,
$\dfrac{64-52}{5}=K[\dfrac{64+52}{2}-\theta_0] \dfrac{12}{5}=K[\dfrac{116}{2}-\theta_0] 2.4=K[58-\theta_0]$ --------(2)
Dividing (1) by (2)
$\dfrac{3.2}{2.4}=K[72-θ0]K[58-θ0]$ (K will get cancel on numerical and denominator)
We can further reduce the equation further by cross multiplication,
$3.2(58-\theta_0)=2.4(72-\theta_0)$
$\implies 185.6-3.2\theta_0=172.8-{2.4\theta}_0$
$\implies 185.6-172.8=3.2\theta_0-{2.4\theta}_0$
$\implies 12.8=0.8\theta_0$
$\implies \theta_0=\dfrac{12.8}{0.8}$
$\therefore \theta_0$ = $16^0C.$

Hence, the correct answer is option (B)

Note:
Newton’s law of cooling describes the cooling of a warmer object to the cooler temperature of the environment. Here, the constant ‘k’ depends on the property of the material which is cooled. It does not depend on any other quantities in the equation discussed below.