A cup of coffee cools from $90^\circ\text{C}$ to $80^\circ\text{C}$ in $t$ minutes, when the room temperature is $20^\circ\text{C}$. The time taken by a similar cup of coffee to cool from $80^\circ\text{C}$ to $60^\circ\text{C}$ at a room temperature same at $20^\circ\text{C}$ is: (2021)
Solution:
Using average form of Newton's law of cooling: $\frac{90-80}{t} = K(\frac{90+80}{2}-20) \Rightarrow \frac{10}{t} = K(65)$. For second case: $\frac{80-60}{t'} = K(\frac{80+60}{2}-20) \Rightarrow \frac{20}{t'} = K(50)$. Dividing the two equations yields $t' = \frac{13}{5}t$.
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