Question

A body cools from \( 80^{\circ}C \) to \( 50^{\circ}C \) in 5 min. Calculate the time it takes to cool from \( 60^{\circ}C \) to \( 30^{\circ}C \) if the surrounding temperature is \( 20^{\circ}C \)

• A. 5 min
• B. 8 min
• C. 9 min
• D. 6 min

Hint:

Using the average temperature \( \frac{\theta_1 + \theta_2}{2} \) as an approximation for the body's temperature during the interval is a highly efficient way to solve cooling problems without using calculus integration.

Answer 1

The Correct Option is C

Concept: According to Newton's Law of Cooling, the rate of change of temperature is directly proportional to the temperature difference between the body and its surroundings: \[ \frac{\theta_1 - \theta_2}{t} = K \left[ \frac{\theta_1 + \theta_2}{2} - \theta_s \right] \]

Step 1: Applying the cooling formula to the first interval.
The body cools from \( 80^{\circ}C \) to \( 50^{\circ}C \) in \( 5 \, \text{min} \) with surrounding temperature \( 20^{\circ}C \): \[ \frac{80 - 50}{5} = K \left[ \frac{80 + 50}{2} - 20 \right] \] \[ \frac{30}{5} = K [65 - 20] \implies 6 = 45K \implies K = \frac{6}{45} = \frac{2}{15} \]

Step 2: Applying the cooling formula to the second interval.
The body cools from \( 60^{\circ}C \) to \( 30^{\circ}C \) over an unknown time \( t \): \[ \frac{60 - 30}{t} = K \left[ \frac{60 + 30}{2} - 20 \right] \] \[ \frac{30}{t} = K [45 - 20] \implies \frac{30}{t} = 25K \]

Step 3: Substituting the value of constant K to solve for \( t \).
Substitute \( K = \frac{2}{15} \): \[ \frac{30}{t} = 25 \left(\frac{2}{15}\right) = \frac{50}{15} = \frac{10}{3} \] \[ 10t = 90 \implies t = 9 \, \text{min} \] This matches option (C) perfectly.