Question

A cup of tea cools from ${{80}^{\circ }}C$ to ${{60}^{\circ }}C$ in 1 min. The ambient temperature is ${{30}^{\circ }}C$. In cooling from ${{60}^{\circ }}C$ to ${{50}^{\circ }}C$, it will take:
A: 50 sec
B: 98 sec
C: 60 sec
D: 48 sec

Answer 1

We know that when a hot cup of tea is left in a room, it gets eventually cooler till the temperature of the cup gets similar to that of the temperature of the air that surrounds the cup. This is because the decrease in internal energy of the cup of tea is equal to the increase in internal energy of the surrounding air.

Formulas used:
$\dfrac{{{T}_{1}}-{{T}_{2}}}{t}=K\left( \dfrac{{{T}_{1}}+{{T}_{2}}}{2}-{{T}_{0}} \right)$
Where ${{T}_{1}}$ is the initial temperature, ${{T}_{2}}$ is the final temperature, ${{T}_{0}}$ is the ambient temperature and K is the constant.

Complete step by step answer:
Newton’s law of cooling states that the rate at which the object’s temperature decreases is directly proportional to the difference between the temperature of the object and the ambient temperature.
In the question, we are given two cases.
In the first case, the initial temperature is ${{80}^{\circ }}C$and the final temperature is ${{60}^{\circ }}C$.
The ambient temperature is ${{30}^{\circ }}C$and the time taken is 60seconds.
We know that $\dfrac{{{T}_{1}}-{{T}_{2}}}{t}=K\left( \dfrac{{{T}_{1}}+{{T}_{2}}}{2}-{{T}_{0}} \right)$
Upon substitution, we obtain
$\begin{aligned} & \dfrac{80-60}{60}=K\left( \dfrac{80+60}{2}-30 \right) \\\ & \Rightarrow \dfrac{1}{3}=40K \\\ & \therefore K=\dfrac{1}{120} \\\ \end{aligned}$
In second case, initial temperature is ${{60}^{\circ }}C$and final temperature is ${{50}^{\circ }}C$
On applying above formula,
$\begin{aligned} & \dfrac{60-50}{t}=\dfrac{1}{120}\left( \dfrac{60+50}{2}-30 \right) \\\ & \Rightarrow \dfrac{10}{t}=\dfrac{1}{120}\left( \dfrac{60+50}{2}-30 \right) \\\ & \Rightarrow \dfrac{10}{t}=\dfrac{25}{120} \\\ & \therefore t=\dfrac{1200}{25}=48\sec \\\ & \\\ \end{aligned}$

Hence, in cooling from ${{60}^{\circ }}C$ to ${{50}^{\circ }}C$, it will take 48seconds.
Therefore D is the correct answer.

Note: Newton’s law is useful to study water cooling as we can find the cooling rate of water in hot pipes using this law. It can tell us how fast a water heater gets cooled when we turn off the breaker.