Question

A body cools from \(80^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in 5 min . In the next time of ' t ' in, the body continues to cool from \(50^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\). The total time taken by the body to cool from \(80^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\) is [The temperature of the surroundings is \(20^{\circ} \mathrm{C}\).]

• A. 10 min
• B. 7.5 min
• C. 15.0 min
• D. 12.5 min

Hint:

Use the average-temperature form of Newton's law for each interval to find the cooling constant, then the second time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Newton's law: \(\dfrac{T_1-T_2}{t}=k\left(\dfrac{T_1+T_2}{2}-T_s\right)\), with \(T_s=20^{\circ}\text{C}\).

Step 2: First interval
\(\dfrac{80-50}{5}=k\left(65-20\right)\)
\(6=45k\), so \(k=\dfrac{2}{15}\ \text{min}^{-1}\).

Step 3: Second interval
\(\dfrac{50-30}{t}=k(40-20)\)
\(\dfrac{20}{t}=\dfrac{2}{15}\times20\), so \(t=7.5\ \text{min}\).

Step 4: Total time
Total \(=5+7.5=12.5\ \text{min}\).

Final Answer:
The total time to cool from 80 to 30 degrees Celsius is 12.5 minutes. \[ \boxed{\text{(D) }12.5\ \text{min}} \]