Question

An ordinary body cools from \(4θ\) to \(3θ\) in 't' minutes. The temperature of the body after next 't' minutes is (Assume Newton's law of cooling and room temperature as \(θ\))

• A. \(\frac{2θ}{3}\)
• B. \(\frac{7θ}{3}\)
• C. \(\frac{5θ}{3}\)
• D. \(\frac{8θ}{3}\)

Hint:

The excess temperature over the room falls by the same factor each interval.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Newton's law of cooling says the rate of fall of temperature is proportional to the excess temperature over the surroundings. This leads to an exponential decay of the excess temperature.

Step 2: Excess temperatures:
Room temperature is \(\theta\). At the start the excess is \(4\theta-\theta=3\theta\). After \(t\) minutes it is \(3\theta-\theta=2\theta\).

Step 3: Decay factor:
In every interval of \(t\) minutes, the excess is multiplied by the same factor: \(\dfrac{2\theta}{3\theta}=\dfrac23\).

Step 4: Next interval:
After another \(t\) minutes, the excess is \(2\theta\times\dfrac23=\dfrac{4\theta}3\).

Step 5: Temperature:
\[ T=\theta+\frac{4\theta}3=\frac{7\theta}3 \]
Option (B).

Final Answer:
The temperature becomes 7 theta / 3. \[ \boxed{\frac{7\theta}{3}} \]