A cup of coffee cools from $90^\circ C$ to $80^\circ C$ in t minutes when the room temperature is $20^\circ C$. The time taken by the similar cup of coffee to cool from $80^\circ C$ to $60^\circ C$ at the same room temperature is :

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YESPYQ · Accepted Answer

Correct answer: d) $\frac{13}{5}t$. From Newton's law of cooling $\frac{90-80}{t}=k\left(\frac{90+80}{2}-20\right)\ \frac{80-60}{{t}^{'}}=k\left(\frac{80+60}{2}-20\right)$ Dividing (i) by (ii) $\frac{10\times {t}^{'}50}{t\times 20}=65\Rightarrow {t}^{'}=\frac{13}{5}t$

A cup of coffee cools from \(90^\circ C\) to \(80^\circ C\) in t minutes when the room temperature is \(20^\circ C\). Th…

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