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
$-\frac{d Q}{d t}=K\left(T-T_{0}\right)$, 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
$t^{\prime}=\frac{50}{30} \min =\frac{5}{3} \min =1 \min 40 s$