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A body cools from \[{60^o}C\] to ${50^o}C$ in 5 minutes. If the temperature of the surrounding is ${30^o}C$ calculate the time period required by the body to cool to ${40^o}C$ .
(A) 8.33 min
(B) 5 min
(C) 8.9 min
(D) Less than 5 min

Last updated date: 22nd Feb 2024
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IVSAT 2024
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Hint Apply Newton’s Law of Cooling and find out the value of temperature changing coefficient for the first case of cooling. Newton’s law of cooling is used to calculate the value of the rate of change of temperature. Now use this value of temperature changing coefficient to find out the value of the time required for the body to cool down in the second case.
Formula used: $\dfrac{{dT}}{{dt}} = K\left[ {\dfrac{{\left( {{T_i} + {T_f}} \right)}}{2} - {T_o}} \right]$

Complete Step by step solution
Here we will use Newton’s Law of Cooling,
where rate of cooling, $\dfrac{{dT}}{{dt}} = K\left[ {\dfrac{{\left( {{T_i} + {T_f}} \right)}}{2} - {T_o}} \right]$.
Here,$t$ is the time taken,
$K$is the temperature changing coefficient,
${T_i}$ is the initial temperature,
${T_f}$ is the final temperature, and
${T_o}$ is the temperature of the surroundings.
For the first case,
${T_i} = {60^o}C,{T_f} = {50^o}C,{T_o} = {30^o}C,dt = 5\min $
Let us substitute these values in the above equation.
$\therefore \dfrac{{\left( {60 - 50} \right)}}{5} = K\left[ {\dfrac{{60 + 50}}{2} - 30} \right]$
$ \Rightarrow \dfrac{{10}}{5} = K\left[ {\dfrac{{110}}{2} - 30} \right]$
$ \therefore 2 = K(55 - 30)$
On further simplifying the equation we get,
$ \Rightarrow 2 = K25$
$ \therefore K = \dfrac{2}{{25}}$
In the second case we have the values as:
Initial temperature, ${T_i} = {50^o}C,$
Final temperature, ${T_f} = {40^o}C,$
Surrounding temperature, ${T_0} = {30^o}C,$
Temperature changing coefficient as found out from the first case, $K = \dfrac{2}{{25}}$
Substituting these values in the equation of Newton’s Law of Cooling we get,
$\dfrac{{\left( {50 - 40} \right)}}{{dt}} = \dfrac{2}{{25}}\left[ {\dfrac{{\left( {50 + 40} \right)}}{2} - 30} \right]$
$ \Rightarrow \dfrac{{10}}{{dt}} = \dfrac{2}{{25}}\left( {\dfrac{{90}}{2} - 30} \right)$
$ \Rightarrow \dfrac{{10}}{{dt}} = \dfrac{2}{{25}}\left( {45 - 30} \right)$
On further simplifying the equation we get,
$ \Rightarrow \dfrac{{10}}{{dt}} = \dfrac{2}{{25}}\left( {15} \right)$
$ \therefore dt = \dfrac{{25}}{3} = 8.33$
Therefore the time taken for the body to cool down from ${50^o}C$ to ${40^o}C$ , with a surrounding temperature of ${30^o}C$ is 8.33 minutes.

Hence option A. is the correct answer.

We should be careful of the fact that the question mentions the cool down of the body from ${60^o}C$to ${50^o}C$ in the first case, and then further the body is cooled down from ${50^o}C$ to ${40^o}C$. Different ranges of temperature difference would yield different results. The unit of temperature cooling coefficient is ${s^{ - 1}}$ , hence we can see that the time taken is dependent on the temperature cooling coefficient.