A cup of tea cools from \( 65.5^{\circ} \mathrm{C} \) to \( 62.5^{\circ} \mathrm{C} \) in one minute in a room at
\( 22.5^{\circ} \mathrm{C} \). How long will it take to cool from \( 46.5^{\circ} \mathrm{C} \) to \( 40.5^{\circ} \mathrm{c} \) in the same room?

Given, cup of tea cools from \(65.5^{\circ} \mathrm{C}\) to \(62.5^{\circ} \mathrm{C}\) in 1 minute.
Room temperature \(=22.5^{\circ} \mathrm{C}\)
We know that \(\frac{d T}{d t}=k\left(\theta-\theta_{0}\right) \rightarrow(1)\)
For 1st case \(: \frac{d T}{d t}=\frac{65.5-62.5}{1 \mathrm{~min}}=\frac{3^{\circ} \mathrm{C}}{1}\)
Also,
\(\theta-\theta_{0}=\left(\frac{65.5+62.5}{2}\right)-22.5=64-22.5\)
\(\theta-\theta_{0}=41.5^{\circ} \mathrm{C}\)
Now, substituting in Eq. (1), we get
\(3=k(41.5)\)
\(\Rightarrow k=\frac{3}{41.5} \rightarrow(2)\)
For 2 nd case \(: \frac{d T}{d t}=\frac{46.5-40.5}{t}=\frac{6^{\circ} \mathrm{C}}{t}\)
Also,
\(\theta-\theta_{0}=\left(\frac{46.5+40.5}{2}\right)-22.5=43.5-22.5\)
\(\Rightarrow \theta-\theta_{0}=21^{\circ} \mathrm{C}\)
Again, substituting in Eq. (1), we get
\(\frac{6}{t}=k \times 21\)
Substitute value of k from Eq. (2), we get
\(\frac{6}{t}=\frac{3}{41.5} \times 21\)
\(\Rightarrow t=\frac{6 \times 41.5}{3 \times 21}=3.9523\)