Question

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

• A. 3.5
• B. 14
• C. 7
• D. 10

Hint:

Rate of cooling is proportional to the temperature difference between the body and surroundings.

Answer 1

The Correct Option is C

To solve this problem, we need to apply Newton's Law of Cooling, which is mathematically represented as:

\[\frac{dT}{dt} = -k (T - T_s)\]

where:

• \(T\) is the temperature of the object at time \(t\).
• \(T_s\) is the surrounding temperature.
• \(k\) is a constant.

The solution to this differential equation provides the formula:

\(T(t) = T_s + (T_0 - T_s) e^{-kt}\)

Using Newton's law, we know:

1. The body cools from \(60^\circ C\) to \(40^\circ C\) in 7 minutes. Let's find the constant \(k\).
2. Initial temperature, \(T_0 = 60^\circ C\), \(T(7) = 40^\circ C\), and surrounding temperature, \(T_s = 10^\circ C\).

Substitute in the formula:

\(40 = 10 + (60 - 10) e^{-7k}\)

Simplifying, we get:

\(30 = 50 e^{-7k}\)

or

\(e^{-7k} = \frac{3}{5}\)

Taking logarithm on both sides:

\(-7k = \ln\left(\frac{3}{5}\right)\)

Therefore,

\(k = -\frac{1}{7} \ln\left(\frac{3}{5}\right)\)

1. Now, using the same law of cooling, let's find the time taken to cool from \(40^\circ C\) to \(28^\circ C\).
2. Initial temperature now is \(T_0 = 40^\circ C\) and \(T(t) = 28^\circ C\).

Substitute in the formula:

\(28 = 10 + (40 - 10) e^{-kt}\)

This simplifies to:

\(18 = 30 e^{-kt}\)

or

\(e^{-kt} = \frac{3}{5}\)

We know \(e^{-7k} = \frac{3}{5}\) and here \(e^{-kt} = e^{-7k}\). Comparing both, \(t = 7 \text{ min}\).

Therefore, the time taken for the body to cool from \(40^\circ C\) to \(28^\circ C\) is 7 minutes.