Question

Ram wants to drink tea, when it is at $50{\rm{^\circ C}}$. He ordered tea which arrived at a temperature of $80{\rm{^\circ C}}$. Tea takes one minute to cool from $80{\rm{^\circ C}}$ to $60{\rm{^\circ C}}$. If the room temperature is $30{\rm{^\circ C}}$, how long does he have to wait to drink the tea?
(A) less than one minute
(B) two more minutes
(C) half a minute
(D) nearly three minute

Answer 1

The law of cooling provides the information about the loss of heat in unit second and according to Newton’s Law of cooling, the rate of cooling of a matter is directly proportional to the difference in mean temperature of the matter and the temperature of the surroundings. The mathematical expression for the Newton’s law of cooling is given as follows,
$\dfrac{{dT}}{t} \propto \left( {{T_m} - {T_s}} \right)$
Here, $dT$ is a change in temperature in time $t$, ${T_m}$ is the mean temperature of the matter and ${T_s}$ is the mean temperature of the surrounding.

Complete step-by-step solution:
The initial temperature of the tea is: ${T_1} = 80{\rm{^\circ C}}$
It is given that tea take $t = 1\;{\rm{min}}$ to cool from ${T_1} = 80{\rm{^\circ C}}$ to ${T_2} = 60{\rm{^\circ C}}$
The final desired temperature of the tea after cooling is: ${T_3} = 50^\circ C$
the temperature of the room is: ${T_s} = 30{\rm{^\circ C}}$
the expression for the mean temperature is given as follows.
${T_m} = \dfrac{{{T_1} + {T_2}}}{2}$
Now, we will apply Newton’s law of cooling for temperature ${T_1}$ and ${T_2}$ .
$\begin{array}{l} \dfrac{{dT}}{t} \propto \left( {{T_m} - {T_s}} \right)\\\ \Rightarrow \dfrac{{dT}}{t} = k\left( {\dfrac{{{T_1} + {T_2}}}{2} - {T_s}} \right)\\\ \Rightarrow \dfrac{{{T_2} - {T_1}}}{t} = k\left( {\dfrac{{{T_1} + {T_2}}}{2} - {T_s}} \right) \end{array}$
Here, $k$ is the constant.
Substitute all the values in the above expression.
$\begin{array}{l} \Rightarrow \dfrac{{80 - 60}}{1} = k\left( {\dfrac{{80 + 60}}{2} - 30} \right)\\\ \Rightarrow \dfrac{{20}}{1} = k\left( {70 - 30} \right)\\\ \Rightarrow k = \dfrac{1}{2} \end{array}$
Here, we have the value of $k$ is $\dfrac{1}{2}$ .
Now we apply Newton's law of cooling for the temperature ${T_2}$ and the desired temperature of the tea ${T_3}$ .
$\begin{array}{l} \Rightarrow \dfrac{{dT}}{t} = k\left( {\dfrac{{{T_1} + {T_2}}}{2} - {T_s}} \right)\\\ \Rightarrow \dfrac{{{T_2} - {T_3}}}{t} = k\left( {\dfrac{{{T_3} + {T_2}}}{2} - {T_s}} \right) \end{array}$
Substitute all the values in the above expression.

\Rightarrow \dfrac{{60 - 50}}{t} = k\left( {\dfrac{{60 + 50}}{2} - 30} \right)\\\ \Rightarrow \dfrac{{10}}{t} = k\left( {55 - 30} \right) \end{array}$$ Substitute the value of the constant $k = \dfrac{1}{2}$ . $$\begin{array}{l} \Rightarrow \dfrac{{10}}{t} = \dfrac{1}{2}\left( {55 - 30} \right)\\\ \Rightarrow t = \dfrac{{20}}{{25}}\;{\rm{min}}\\\ \Rightarrow t = 0.8\;\min \end{array}$$ The tea takes $0.8\;{\rm{min}}$ to cool from $60{\rm{^\circ C}}$ to ${\rm{50^\circ C}}$ . Therefore, Ram has to wait for less than one minute to drink the tea and **the correct answer is option (A).** **Note:** In these types of questions they always calculate the cooling constant and always take all temperature in the same units. Newton's law of cooling can be used to determine the initial and the final temperature of the matter.