Question

A hot body is allowed to cool. The surrounding temperature is constant at 30 ${}^\circ C$. The body takes time ${{t}_{1}}$to cool from 90${}^\circ C$ to 89 ${}^\circ C$ and time ${{t}_{2}}$ to cool from 60${}^\circ C$ to 59.5${}^\circ C$ . Then,
A-${{t}_{2}}={{t}_{1}}$
B-${{t}_{2}}=\dfrac{{{t}_{1}}}{2}$
C-${{t}_{2}}=4{{t}_{1}}$
D-${{t}_{2}}={{t}_{1}}$

Answer 1

This is a problem of Newton’s law of cooling as we have to determine the rate of cooling. We are given two conditions in which a body cools down from a certain temperature to another, We need to determine the relationship between the two times.

Complete step by step answer:
According to Newton's Law of Cooling, rate of heat transfer is given by
$\dfrac{dQ}{dt}=kA(T-{{T}_{0}})$
The surrounding temperature is constant at 30 ${}^\circ C$. The body takes time ${{t}_{1}}$to cool from 90${}^\circ C$ to 89 ${}^\circ C$writing this mathematically,
$\dfrac{d{{Q}_{1}}}{dt}=60kA$------(1)
For the second case,
$\dfrac{d{{Q}_{2}}}{dt}=kA(90-60)$
$\implies$ $\dfrac{d{{Q}_{2}}}{dt}=30kA$------(2)
But heat is given as $Q=mc\Delta T$
In the first case, the difference of temperature is (90-89)= 1 ${}^\circ C$and for the second case, it is (60-59.5)=0.5 ${}^\circ C$
$\Rightarrow \dfrac{\Delta {{Q}_{2}}}{\Delta {{Q}_{1}}}=\dfrac{mc\times .5}{mc\times 1}$
$\Rightarrow \dfrac{\Delta {{Q}_{2}}}{\Delta {{Q}_{1}}}=\dfrac{mc\times .5}{mc\times 1}=0.5$-------(3)
From (1), (2) and (3)
${{t}_{2}}={{t}_{1}}$

So, the correct answer is “Option A”.

Note:
Newton’s law of cooling tells us the rate at which a body exposed to atmosphere changes temperature through loss of heat to the surrounding which is approximately proportional to the difference between the object’s temperature and its surroundings, provided the difference is small. As per this law rate of cooling is dependent upon the temperature, suppose we want a hot cup of tea to cool down, a cup of hot coffee will cool more quickly if we put it in the refrigerator.