Question

A hot body is placed in cold surroundings. Its rate of cooling is ${3^0}C$ per minute when its temperature is ${70^0}C$ and ${1.5^0}C$ per minute when its temperature is ${50^0}C$. Its rate of cooling when temperature is ${40^0}C$ is
(a){\text{ 0}}{\text{.2}}{{\text{5}}^0}C/\min \\\ (b){\text{ 0}}{\text{.}}{{\text{5}}^0}C/\min \\\ (c){\text{ 0}}{\text{.7}}{{\text{5}}^0}C/\min \\\ (d){\text{ }}{{\text{1}}^0}C/\min \\\

Answer 1

Hint – In this question write the rate of cooling as $\dfrac{{d\theta }}{{dt}}$ where $\theta$ is the temperature, then use the concept that rate of cooling is written as $\dfrac{{d\theta }}{{dt}} = k\left( {\theta - {\theta _0}} \right)$, then formulate equations to get the value of k and ${\theta _0}$. Using the obtained value find the rate of cooling for ${40^0}C$.
Formula used – $\dfrac{{d\theta }}{{dt}} = k\left( {\theta - {\theta _0}} \right)$.

Complete step-by-step solution -
Given data:
When $\theta = {70^0}C$, its rate of cooling, $\dfrac{{d\theta }}{{dt}} = {3^0}C$/min.
When $\theta = {50^0}C$, its rate of cooling, $\dfrac{{d\theta }}{{dt}} = {1.5^0}C$/min.
Now we have to find out rate of cooling when $\theta = {40^0}C$
As we know that rate of cooling is given as
$\Rightarrow \dfrac{{d\theta }}{{dt}} = k\left( {\theta - {\theta _0}} \right)$, where k = constant and ${\theta _0}$ = room temperature.
Now from first condition we have,
$\Rightarrow 3 = k\left( {{{70}^0} - {\theta _0}} \right)$...................... (1)
Now from second condition we have,
$\Rightarrow 1.5 = k\left( {{{50}^0} - {\theta _0}} \right)$................... (2)
Now divide these two equations we have,
$\Rightarrow \dfrac{3}{{1.5}} = \dfrac{{k\left( {{{70}^0} - {\theta _0}} \right)}}{{k\left( {{{50}^0} - {\theta _0}} \right)}}$
Now simplify this we have,
$\Rightarrow 2\left( {{{50}^0} - {\theta _0}} \right) = \left( {{{70}^0} - {\theta _0}} \right)$
$\Rightarrow {100^0} - {70^0} = 2{\theta _0} - {\theta _0}$
$\Rightarrow {30^0} = {\theta _0}$
Now from equation (1) we have,
$\Rightarrow 3 = k\left( {{{70}^0} - {{30}^0}} \right) = k\times {40^0}$
$\Rightarrow k = \dfrac{3}{{40}}$
So rate of cooling when $\theta = {40^0}C$
$\Rightarrow \dfrac{{d\theta }}{{dt}} = \dfrac{3}{{40}}\left( {{{40}^0} - {{30}^0}} \right) = \dfrac{3}{4} = {0.75^0}C$/min.
So this is the required answer.
Hence option (C) is the correct answer.

Note – Newton 's cooling law is a very important concept when dealing with issues of this nature. It states that the rate of temperature change for an object or any material is directly proportional to the difference in temperature between the surrounding temperature and the temperature of the body.