Question

A cup of tea cools from $81^{\circ }C$ to $79^{\circ }C$ in $1\min$. The ambient temperature is $30^{\circ }C$. What time is needed for cooling of same cup of tea in same ambience from $61^{\circ }C$ to $59^{\circ }C$ ?

• A. 1 min 40 s
• B. 1 min 6 s
• C. 1 min 22 s
• D. 4 min 3 s

Answer 1

Answer: (A)

Given, in first case, $T_1=81^{\circ }C,T_2=79^{\circ }C$,
$T_0=30^{\circ }C$ and $t=1\min$.
As fall in temperature in accordance with Newton's law of cooling expression is
$-\frac{dQ}{dt}=K\left(T-T_0\right)$, we can write
$\Rightarrow \left(\frac{T_1-T_2}{t}\right)=-K\left(\frac{T_1+T_2}{2}-T_0\right)$
$\Rightarrow \frac{81-79}{1\min }=-K\left(\frac{81+79}{2}-30\right)$
$\Rightarrow \frac{2}{1\min }=-K\times 50$
and in second case, $T_1^{\prime }=61^{\circ }C,T_2^{\prime }=59^{\circ }C$.
If time of cooling be $t^{\prime }$, then
$\frac{61-59}{t^{\prime }}=-K\left[\frac{61+59}{2}-30\right]$
or $\frac{2}{t^{\prime }}=-K\times 30$
On dividing Eq. (i) by Eq. (ii), we get
$t^{\prime }=\frac{50}{30}\min =\frac{5}{3}\min =1\min 40s$