Question

A metal sphere cools from \(80^{\circ }\text{C}\) to \(60^{\circ }\text{C}\) in 10 minute. The time taken (in minute) by it to cool from \(60^{\circ }\text{C}\) to \(40^{\circ }\text{C}\) if the room temperature is \(30^{\circ }\text{C}\), is

• A. \(10\)
• B. \(20\)
• C. \(30\)
• D. \(40\)

Hint:

Use the average form of Newton's law: fall in temperature over time equals a constant times the average excess temperature.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Newton's law (average form): \(\dfrac{T_{1}-T_{2}}{t}=k\left(\dfrac{T_{1}+T_{2}}{2}-T_{0}\right)\).

Step 2: First interval
\(\dfrac{80-60}{10}=k(70-30)\)
\(2=40k\), so \(k=0.05\).

Step 3: Second interval
\(\dfrac{60-40}{t}=0.05(50-30)\)
\(\dfrac{20}{t}=1\), so \(t=20\) min.

Final Answer:
It takes 20 minutes to cool from \(60^{\circ}\text{C}\) to \(40^{\circ}\text{C}\). \[ \boxed{\text{(B) }20} \]