Question

If an object cools down from 80°C to 60°C in 5 min in a surrounding of temperature 20°C. The time taken to cool from 60°C to 40°C will be (Assume Newton’s law of cooling to be valid)

• A. \(\frac{25}{3}\) min
• B. 5 min
• C. \(\frac{25}{4}\) min
• D. 5 min

Answer 1

The Correct Option is A

To determine the time taken for an object to cool from 60°C to 40°C using Newton's law of cooling, we first need to understand the formula:

\(T(t) = T_s + (T_i - T_s) e^{-kt}\)

Here:

• \(T(t)\) is the temperature of the object at time \(t\).
• \(T_s\) is the surrounding temperature, 20°C in this case.
• \(T_i\) is the initial temperature of the object.
• \(k\) is a constant for a particular situation.

First, find the constant \(k\). From the given question, the object cools down from 80°C to 60°C in 5 minutes. Thus:

1. From \(80^\circ C\) to \(60^\circ C\) in the surroundings of \(20^\circ C\):
   \(60 = 20 + (80 - 20) e^{-5k}\)
2. Simplifying:
   \(40 = 60 e^{-5k} \Rightarrow e^{-5k} = \frac{2}{3}\)
3. Therefore, \(-5k = \ln \left( \frac{2}{3} \right) \Rightarrow k = -\frac{1}{5} \ln \left( \frac{2}{3} \right)\)

Now, find the time taken to cool from \(60^\circ C\) to \(40^\circ C\):

1. Equation Setup:
   \(40 = 20 + (60 - 20) e^{-kt}\)
2. Simplifying:
   \(20 = 40 e^{-kt} \Rightarrow e^{-kt} = \frac{1}{2}\)
3. Using the previous \(k\) value:
   \(-kt = \ln \left( \frac{1}{2} \right)\)
4. Substitute \(k\):
   \(t = \frac{\ln \left( \frac{1}{2} \right)}{-\frac{1}{5} \ln \left( \frac{2}{3} \right)}\)
5. Simplifying further:
   Using the property \(\ln(2) = - \ln\left(\frac{1}{2}\right)\), find that
   \(t = \frac{5 \ln \left( \frac{2}{3} \right)}{\ln \left( \frac{1}{2} \right)}\)
6. This yields approximately:
   \(t = \frac{25}{3} \, \text{minutes}\)

Thus, the time taken for the object to cool from 60°C to 40°C is \(\frac{25}{3}\) minutes.