Question

A body cools in \(7\) minutes from \(60^{\circ}\)C to \(40^{\circ}\)C. What time (in minutes) does it take to cool from \(40^{\circ}\)C to \(28^{\circ}\)C, if its surrounding temperature is \(10^{\circ}\)C ? (Newton's law of cooling holds good)

• A. \(3.5\)
• B. \(10\)
• C. \(11\)
• D. \(7\)

Hint:

Use the average-temperature form of Newton's law for each step.

Answer 1

The Correct Option is D

Step 1: Use the exponential form:
The excess temperature falls as $\theta=\theta_0e^{-kt}$. First step: excess falls from 50 to 30 in 7 min.

Step 2: Find k:
$\dfrac{30}{50}=e^{-7k}$, so $7k=\ln\dfrac53$ and $k=0.0730$ per min.

Step 3: Second step:
Excess goes from 30 to 18: $\dfrac{18}{30}=e^{-kt}$, so $kt=\ln\dfrac53$. Hence $t=7$ min exactly, since $\dfrac{18}{30}=\dfrac{30}{50}=0.6$. Option D.

Final Answer:
The excess temperature falls by the same ratio 0.6 in both steps, so the time is 7 min. \[ \boxed{\text{(D) }7} \]