Question

A body cools from \(100^{\circ }\)C to \(80^{\circ }\)C in 5 minutes in air at \(20^{\circ }\)C. The time taken to cool from \(80^{\circ }\)C to \(60^{\circ }\)C will be:

• A. Less than 5 minutes
• B. Equal to 5 minutes
• C. Greater than 5 minutes
• D. Cannot be determined

Hint:

By Newton's law the temperature excess over air decays exponentially; compare the ratio of excess temperatures in each interval.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Newton's law of cooling: \(\frac{d\theta}{dt}=-k\theta\), where \(\theta\) is the excess temperature over the air at \(20^{\circ}\)C. So \(t=\frac{1}{k}\ln\frac{\theta_{1}}{\theta_{2}}\).

Step 2: First interval
Excess goes from \(80^{\circ}\) to \(60^{\circ}\) in 5 minutes: \(5=\frac{1}{k}\ln\frac{80}{60}=\frac{1}{k}\ln\frac{4}{3}\).

Step 3: Second interval
Excess goes from \(60^{\circ}\) to \(40^{\circ}\): \(t=\frac{1}{k}\ln\frac{60}{40}=\frac{1}{k}\ln\frac{3}{2}\).

Step 4: Compare
\(\ln\frac{3}{2}\approx 0.405\) is larger than \(\ln\frac{4}{3}\approx 0.288\).
So \(t=5\times\frac{0.405}{0.288}\approx 7\) minutes.

Final Answer:
The second cooling takes about 7 minutes, which is more than 5 minutes. \[ \boxed{\text{(C) }\text{Greater than 5 minutes}} \]