Question

A body cools from ${62^0}C$ and ${50^0}C$ in \[10{\text{ }}minutes\]. How long will it take to cool to ${42^0}C$, if room temperature is ${26^0}C$ ?
A. $5\min $
B. $7.5\min $
C. $10\min $
D. $12.5\min $

Answer 1

Hint: This problem can be solved by good knowledge of Newton's law of cooling. Newton’s law of cooling is the rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.

Formula used:
Newton’s law of cooling :-
The rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.
Let \[\theta \] and ${\theta _0}$ be the temperature of a body and its surrounding respectively . Let $\dfrac{{dT}}{{dt}}$ be the rate of loss of heat , so from Newton’s
Law of cooling
$\dfrac{{dT}}{{dt}} \propto (\theta - {\theta _0})$
$\dfrac{{dT}}{{dt}} = k(\theta - {\theta _0})$
$\dfrac{{_{{T_i} - {T_f}}}}{t} = k(T - {T_2})$

Complete step by step answer:
- Newton’s law of cooling : The rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.

Time-\[10{\text{ }}min\]
time - ?

Here, Initial temperature (${T_i}$) = \[62^\circ C\]
Final temperature (${T_f}$) = \[50^\circ C\]
Temperature of the surrounding (${T_o}$) = \[26^\circ C\]
t = \[10{\text{ }}min\]
By Newton's law of cooling Rate of cooling
$\dfrac{{_{{T_i} - {T_f}}}}{t} = k(T - {T_2})$
\[ \;\left( {{\text{ }}dT/dt} \right){\text{ }} = {\text{ }}K\left[ {{\text{ }}\left( {Ti + Tf} \right)/2{\text{ }} - {\text{ }}To} \right] \\
  \left( {{\text{ }}Tf{\text{ }} - {\text{ }}Ti} \right)/t{\text{ }} = {\text{ }}K\left[ {{\text{ }}\left( {62{\text{ }} + {\text{ }}50} \right)/2{\text{ }} - {\text{ }}26} \right] \\
  \;\left( {{\text{ }}62 - 50} \right)/10{\text{ }} = {\text{ }}K\left[ {{\text{ }}56{\text{ }} - {\text{ }}26} \right] \\
\]

In second condition,
Initial temperature (${T_i}$) =\[50^\circ C\] \[\]
Final temperature (${T_f}$) = \[42^\circ C\]
Time taken for cooling is t
By Newton's law of cooling $\dfrac{{_{{T_i} - {T_f}}}}{t} = k(T - {T_2})$
\[\left( {{\text{ }}50 - {\text{ }}42} \right)/t{\text{ }} = {\text{ }}k{\text{ }}\left[ {{\text{ }}\left( {50 + 42} \right)/2{\text{ }} - 26} \right]\]---------------(ii)

From equation (i) divided by (ii)
\[\dfrac{{\left( {{\text{ }}62 - 50} \right)/10{\text{ }}}}{{\left( {{\text{ }}50 - {\text{ }}42} \right)/t{\text{ }}}} = \dfrac{{K\left[ {{\text{ }}56{\text{ }} - {\text{ }}26} \right]}}{{k{\text{ }}\left[ {{\text{ }}\left( {50 + 42} \right)/2{\text{ }} - 26} \right]}}\]
\[\dfrac{{12/10{\text{ }}}}{{8/t{\text{ }}}} = \dfrac{{30}}{{20}}\]
\[t = 10\min \]

So, the correct answer is “Option C”.

Note:
This problem can be solved by good knowledge of Newton's law of cooling. Read the problem carefully and apply the concept related to the problem by the given things. Imagination of problems makes it easy to understand the problem. you should have good command in the concept as well as formula related to the concept.