Question 116 of 126
A cup of coffee cools from to in t minutes when the room temperature is . The time taken by the similar cup of coffee to cool from to at the same room temperature is :
Show answer and solution
Answer: (D)
By using average form of Newton's law of cooling
$\begin{aligned}
& \frac{90-80}{\mathrm{t}}=\mathrm{k}\left(\frac{90+80}{2}-20\right) \\
& \frac{80-60}{\mathrm{t}^{\prime}}=\mathrm{k}\left(\frac{80+60}{2}-20\right)
\end{aligned}$
$\begin{aligned}
& \text { (i)/(ii) } \\
& \frac{10 \times \mathrm{t}^{\prime}}{\mathrm{t} \times 20}=\frac{65}{50} \\
& \mathrm{t}^{\prime}=\frac{65}{50} \times 2 \mathrm{t}=\frac{65}{25} \mathrm{t}=\frac{13}{5} \mathrm{t}
\end{aligned}$
$\begin{aligned}
& \frac{90-80}{\mathrm{t}}=\mathrm{k}\left(\frac{90+80}{2}-20\right) \\
& \frac{80-60}{\mathrm{t}^{\prime}}=\mathrm{k}\left(\frac{80+60}{2}-20\right)
\end{aligned}$
$\begin{aligned}
& \text { (i)/(ii) } \\
& \frac{10 \times \mathrm{t}^{\prime}}{\mathrm{t} \times 20}=\frac{65}{50} \\
& \mathrm{t}^{\prime}=\frac{65}{50} \times 2 \mathrm{t}=\frac{65}{25} \mathrm{t}=\frac{13}{5} \mathrm{t}
\end{aligned}$