Question

A body cools in \(7\) minutes from \(60^{\circ}\)C to \(40^{\circ}\)C. What time (in minutes) does it take to cool from \(40^{\circ}\)C to \(28^{\circ}\)C, if its surrounding temperature is \(10^{\circ}\)C ? (Newton's law of cooling holds good)

• A. \(3.5\)
• B. \(10\)
• C. \(11\)
• D. \(7\)

Hint:

Use the average-temperature form of Newton's law for each step.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Newton's law of cooling in the average form is \(\dfrac{T_1-T_2}{t}=k\left(\dfrac{T_1+T_2}{2}-T_0\right)\).

Step 2: First interval:
From \(60^{\circ}\)C to \(40^{\circ}\)C in 7 min with \(T_0=10^{\circ}\)C: \(\dfrac{20}{7}=k(50-10)=40k\), so \(k=\dfrac{1}{14}\) per min.

Step 3: Second interval:
From \(40^{\circ}\)C to \(28^{\circ}\)C: \(\dfrac{12}{t}=k(34-10)=24k=\dfrac{24}{14}\). So \(t=\dfrac{12\times14}{24}=7\) min.

Step 4: Why the other options are wrong.
3.5 and 10 and 11 come from assuming a constant rate of fall or from a wrong average temperature.

Final Answer:
The time is 7 minutes. \[ \boxed{\text{(D) }7} \]