Question

A body cools from temperature \((3\theta)^\circ\)C to \((2\theta)^\circ\)C in 10 minute. Then it cools from \((2\theta)^\circ\)C to \((\theta_1)^\circ\)C in next 10 minute. The room temperature is \((\theta)^\circ\)C. Assuming that the Newton's law of cooling is applicable the value of \((\theta_1)^\circ\)C is

• A. \(\theta\)
• B. \(\frac{5}{4}\theta\)
• C. \(\frac{4}{3}\theta\)
• D. \(\frac{3}{2}\theta\)

Hint:

Equate the rate of fall of temperature to a constant times the difference from room temperature, using average temperatures.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Newton's law gives \(\dfrac{T_1-T_2}{t}=k\left(\dfrac{T_1+T_2}{2}-T_0\right)\) with room temperature \(T_0=\theta\).

Step 2: First interval
\(\dfrac{3\theta-2\theta}{10}=k\left(\dfrac{5\theta}{2}-\theta\right)\), so \(\dfrac{\theta}{10}=\dfrac{3k\theta}{2}\) and \(k=\dfrac{1}{15}\).

Step 3: Second interval
\(\dfrac{2\theta-\theta_1}{10}=\dfrac{1}{15}\left(\dfrac{2\theta+\theta_1}{2}-\theta\right)=\dfrac{\theta_1}{30}\).

Step 4: Calculation
\(3(2\theta-\theta_1)=\theta_1\), so \(6\theta=4\theta_1\) and \(\theta_1=\dfrac{3}{2}\theta\).

Final Answer:
The body cools to \(\tfrac{3}{2}\theta\). \[ \boxed{\text{(D) }\dfrac{3}{2}\theta} \]