Question

The surrounding temperature is \( 20^\circ\text{C} \). A body cools from \( 100^\circ\text{C} \) to \( 60^\circ\text{C} \) in 20 minutes. Find the time taken for the body to cool down to \( 30^\circ\text{C} \).

• A. \( 40 \) minutes
• B. \( 50 \) minutes
• C. \( 60 \) minutes
• D. \( 80 \) minutes

Hint:

In Newton's Law of Cooling, always subtract the surrounding temperature first. The cooling process depends on the temperature difference, not the actual temperature itself.

Answer 1

The Correct Option is C

Concept: Newton's Law of Cooling states that the rate of change of temperature of a body is proportional to the difference between the body temperature and the surrounding temperature. Mathematically: \[ \frac{dT}{dt}=-k(T-T_s) \] Its solution is: \[ \ln\left(\frac{T-T_s}{T_0-T_s}\right)=-kt \] where:
• \(T_s\) = surrounding temperature
• \(T_0\) = initial temperature
• \(T\) = temperature after time \(t\)
• \(k\) = cooling constant

Step 1: Finding the cooling constant \(k\).
Given: \[ T_s=20^\circ C \] Initial temperature: \[ T_0=100^\circ C \] After \(20\) minutes: \[ T=60^\circ C \] Using: \[ \ln\left(\frac{T-T_s}{T_0-T_s}\right)=-kt \] Substitute values: \[ \ln\left(\frac{60-20}{100-20}\right)=-20k \] Simplify: \[ \ln\left(\frac{40}{80}\right)=-20k \] \[ \ln\left(\frac12\right)=-20k \] Using: \[ \ln\left(\frac12\right)=-\ln2 \] Thus: \[ -\ln2=-20k \] Hence: \[ k=\frac{\ln2}{20} \]

Step 2: Finding the total time for cooling to \(30^\circ C\).
Now: \[ T=30^\circ C \] Using the same formula: \[ \ln\left(\frac{30-20}{100-20}\right)=-kt \] Substitute: \[ \ln\left(\frac{10}{80}\right)=-kt \] \[ \ln\left(\frac18\right)=-kt \] Since: \[ \frac18=2^{-3} \] therefore: \[ \ln\left(\frac18\right)=-3\ln2 \] Thus: \[ -3\ln2=-\left(\frac{\ln2}{20}\right)t \] Cancel \(\ln2\): \[ 3=\frac{t}{20} \] Hence: \[ t=60 \text{ minutes} \] Therefore, \[ \boxed{60\text{ minutes}} \]