Question

A cup of tea cools from 80°C to 60°C in 2 minutes when the temperature of the surrounding is 30°C. Time taken to cool from 60° to 50°C is,
A. 1 min 12s
B. 1 min 24s
C. 1 min 36s
D. 1 min 48s

Answer 1

Hint: This question is based on the concept of Newton's law of cooling. It states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surrounding temperature.

Complete step by step answer:
 Using Newton’s law of cooling,
$\left( {\dfrac{{{\theta _1} - {\theta _2}}}{t}} \right) = n\left( {\dfrac{{{\theta _1} - {\theta _2}}}{2} - {\theta _0}} \right)$
Given initial temperature = 30°C
Case1: Temperature falls from 80°C to 60°C,
 $
 \left( {\dfrac{{80 - 60}}{2}} \right) = n\left( {\dfrac{{80 + 60}}{2} - 30} \right) \\
\implies \left( {\dfrac{{20}}{2}} \right) = n\left( {70 - 30} \right) \\
\implies n = \dfrac{1}{4} \\
$
 Case2: Temperature falls from 60° to 50°C,
$
\left( {\dfrac{{60 - 50}}{t}} \right) = n\left( {\dfrac{{60 + 50}}{2} - 30} \right) \\
\implies \left( {\dfrac{{10}}{t}} \right) = n(55 - 30) \\
$
Replacing the value of n=0.25 in the above equation,
$
\left( {\dfrac{{10}}{t}} \right) = \dfrac{1}{4} \times \left( {55 - 30} \right) \\
\implies t = \dfrac{{10 \times 4}}{{(55 - 30)}} = \dfrac{{40}}{{25}} = \dfrac{8}{5} = 1.6\min \\
\implies t = 1\min 36\sec \\
$

So, the correct answer is “Option C”.

Note:
Newton's law of cooling has certain limitations such as the difference in temperature between the body and its surrounding should be less and the heat loss from the body should be in the form of radiation only. The prime limitation of the law is that, during the cooling process of the body, the surrounding temperature must be constant.