Thermal Properties of Matter - Result Question 46
47. An object kept in a large room having air temperature of $25^{\circ} C$ takes 12 minutes to cool from $80^{\circ} C$ to $70^{\circ} C$.
The time taken to cool for the same object from $70^{\circ} C$ to $60^{\circ} C$ would be nearly.
[NEET Odisha 2019]
(a) $15 min$
(b) $10 min$
(c) $12 min$
(d) $20 min$
Show Answer
Answer:
Correct Answer: 47. (a)
Solution:
- (a) According to Newton’s law of cooling
$ \frac{(T_1-T_2)}{t}=K(\frac{T_1+T_2}{2}-T_0) $
$ \begin{align*} & \frac{(80-70)}{12}=K(75-25) \tag{i}\\ & \frac{(70-60)}{t^{\prime}}=K(65-25) \tag{ii} \end{align*} $
$\Rightarrow t^{\prime}=12 \times=15 min$
