Q 11-10-004NEETNEET 2021Top questionMedium
A cup of coffee cools from $90°$C to $80°$C in t minutes, when the room temperature is $20°$C. The time taken by a similar cup of coffee to cool from $80°$C to $60°$C at a room temperature same at $20°$C is :
Answer: (C) $\dfrac{13}{5}t$
Concept. Newton's law of cooling in average form: $\dfrac{\Delta T}{t} = K(T_{\text{avg}} - T_0)$.
Step 1. First cooling ($90 \to 80$).
$$\frac{10}{t} = K(85 - 20) = 65K$$
Step 2. Second cooling ($80 \to 60$).
$$\frac{20}{t'} = K(70 - 20) = 50K$$
Step 3. Divide.
$$\frac{20/t'}{10/t} = \frac{50}{65} \;\Rightarrow\; \frac{2t}{t'} = \frac{10}{13} \;\Rightarrow\; t' = \frac{13}{5}t$$
Answer. $\dfrac{13}{5}t$ (Option C).
Solution by Sreeraj P, M.Sc Physics