Question

A bowl filled with very hot water cools from \(98^{\circ}\text{C}\) to \(86^{\circ}\text{C}\) in 2 minutes when the room temperature is \(22^{\circ}\text{C}\). How long it will take to cool from \(75^{\circ}\text{C}\) to \(69^{\circ}\text{C}\) ?

• A. \(0.5\) minute
• B. \(1.4\) minute
• C. \(1\) minute
• D. \(2\) minute

Hint:

The rate of fall equals k times the excess over room temperature, using the mean temperature.

Answer 1

The Correct Option is B

Step 1: Find k
Fall of \(12^{\circ}C\) in \(2\) min, mean temperature \(92^{\circ}C\), excess \(70^{\circ}C\). Newton's law: \(\frac{12}{2} = k(70)\), so \(k = \frac{3}{35}\ \text{min}^{-1}\).

Step 2: Second interval
Fall of \(6^{\circ}C\), mean temperature \(72^{\circ}C\), excess \(50^{\circ}C\). So \(\frac{6}{t} = k(50)\).

Step 3: Solve
\[ t = \frac{6}{\frac3{35}\times50} = \frac{6\times35}{150} = 1.4\ \text{min} \]
Option (B).

Final Answer:
The time taken is 1.4 minutes. \[ \boxed{\text{(B)}\ 1.4\ \text{minute}} \]