Question

A body cools according to Newton’s law from \(100^\circ\text{C}\) to \(60^\circ\text{C}\) in 20 minutes. The temperature of the surroundings being \(20^\circ\text{C}\), the temperature of the body after one hour is

• A. \(15^\circ\text{C}\)
• B. \(30^\circ\text{C}\)
• C. \(40^\circ\text{C}\)
• D. \(20^\circ\text{C}\)

Hint:

Always subtract the surrounding temperature before applying Newton’s law of cooling.

Answer 1

The Correct Option is B

Step 1: Apply Newton’s law of cooling.
According to Newton’s law, \[ T - T_s = (T_0 - T_s)e^{-kt} \] where \(T_s = 20^\circ\text{C}\).

Step 2: Use the given data to find \(k\).
At \(t = 20\) minutes, \(T = 60^\circ\text{C}\): \[ 60 - 20 = (100 - 20)e^{-20k} \Rightarrow 40 = 80e^{-20k} \Rightarrow e^{-20k} = \frac{1}{2} \]
Step 3: Find temperature after 60 minutes.
\[ T - 20 = 80e^{-60k} = 80\left(\frac{1}{2}\right)^3 = 10 \]
Step 4: Final answer.
\[ T = 20 + 10 = 30^\circ\text{C} \]