Question

For a liquid, it is allowed to cool down in a room whose temperature is $20$, it takes $5\,\min $ to cool down from $50$ to $46$. Find out temperature of the liquid after next $5\,\min $ ?

Answer 1

Hint:In this question, the liquid is allowed to cool in the atmosphere temperature, by using Newton's law of cooling and by using the given temperatures of the atmosphere and the temperature of the liquid and the time taken, the temperature of the liquid for a particular time can be determined.

Useful formula:
Newton’s law of cooling is given by,
$\dfrac{{{T_t} - {T_s}}}{{{T_i} - {T_s}}} = {e^{\dfrac{{ - t}}{i}}}$
Where, ${T_t}$ is the temperature at the time, ${T_s}$ is the temperature of the atmosphere or temperature of the surroundings, ${T_i}$ is the initial temperature of the liquid and $t$ is the time.

Complete step by step solution:
Given that,
The atmospheric temperature or the surrounding temperature is, ${T_s} = 20$
The initial temperature of the liquid is, ${T_i} = 50$
The temperature of the liquid after $5\,\min $ is, ${T_5} = 46$
Now,
Newton’s law of cooling is given by,
$\dfrac{{{T_t} - {T_s}}}{{{T_i} - {T_s}}} = {e^{\dfrac{{ - t}}{i}}}\,...................\left( 1 \right)$
By substituting the temperature and time for the first $5\,\min $ in the equation (1), then the equation (1) is written as,
$\dfrac{{46 - 20}}{{50 - 20}} = {e^{\dfrac{{ - 5}}{i}}}\,......................\left( 2 \right)$
Now the temperature equation for the total time $10\,\min $, is written as,
\[\dfrac{{{T_{10}} - 20}}{{50 - 20}} = {e^{\dfrac{{ - 10}}{i}}}\,.....................\left( 3 \right)\]
The RHS of the equation (3) is also written as,
\[{e^{\dfrac{{ - 10}}{i}}} = {e^{\dfrac{{ - 5}}{i}}} \times {e^{\dfrac{{ - 5}}{i}}}\]
By substituting the values of ${e^{\dfrac{{ - 10}}{i}}}$ and ${e^{\dfrac{{ - 5}}{i}}}$, then
\[\dfrac{{{T_{10}} - 20}}{{50 - 20}} = \left( {\dfrac{{46 - 20}}{{50 - 20}}} \right) \times \left( {\dfrac{{46 - 20}}{{50 - 20}}} \right)\]
On further calculation, then the above equation is written as,
\[\dfrac{{{T_{10}} - 20}}{{30}} = \left( {\dfrac{{26}}{{30}}} \right) \times \left( {\dfrac{{26}}{{30}}} \right)\]
By rearranging the terms, then the above equation is written as,
\[{T_{10}} - 20 = \dfrac{{26 \times 26 \times 30}}{{30 \times 30}}\]
By cancelling the same terms then,
\[{T_{10}} - 20 = \dfrac{{26 \times 26}}{{30}}\]
On multiplying the terms, then
\[{T_{10}} - 20 = \dfrac{{676}}{{30}}\]
On dividing the terms, then the above equation is written as,
\[{T_{10}} - 20 = 22.53\]
By rearranging the terms, then
\[{T_{10}} = 22.53 + 20\]
By adding the terms, then
\[{T_{10}} = 42.53\]
Therefore, after $10\,\min $ the temperature of the liquid is $42.53$.

Note: After the equation (3), the RHS of the equation (3) is split up into two for the further calculation and also for the easy calculation. And then by substituting the values, the temperature of the liquid after next $5\,\min $ can be determined.