Step 1: Understanding the Concept:
This problem is an application of Newton's Law of Cooling, which describes how the temperature of an object changes over time when it is placed in an environment with a different temperature.
The law states that the rate of change of temperature (\( \frac{dT}{dt} \)) is directly proportional to the difference between the object's current temperature (\( T \)) and the constant surrounding temperature (\( T_s \)).
Mathematically, this relationship leads to an exponential decay of the temperature difference over time.
The colder the object gets, the slower it continues to cool because the temperature gradient decreases.
Step 2: Key Formula or Approach:
The differential equation is \( \frac{dT}{dt} = -k(T - T_s) \).
The solution to this equation in terms of logarithms 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
Alternatively, we can use the property that the ratios of temperature differences over equal intervals of time follow a geometric progression.
Step 3: Detailed Explanation:
Given constants: \( T_s = 20^{\circ}\text{C} \).
Initial condition: At \( t = 0 \), \( T_0 = 100^{\circ}\text{C} \).
First cooling stage: At \( t_1 = 20 \) minutes, \( T = 60^{\circ}\text{C} \).
Let's find the constant \( k \) using the first set of data.
Using the formula:
\[ \ln\left(\frac{60 - 20}{100 - 20}\right) = -k(20) \]
\[ \ln\left(\frac{40}{80}\right) = -20k \]
\[ \ln\left(\frac{1}{2}\right) = -20k \]
Since \( \ln(1/2) = -\ln 2 \), we have:
\[ -\ln 2 = -20k \implies k = \frac{\ln 2}{20} \]
Now, we need to find the total time \( t \) when the temperature reaches \( 30^{\circ}\text{C} \).
Using the formula again from the starting point:
\[ \ln\left(\frac{30 - 20}{100 - 20}\right) = -kt \]
\[ \ln\left(\frac{10}{80}\right) = -kt \]
\[ \ln\left(\frac{1}{8}\right) = -kt \]
We know that \( \frac{1}{8} = \left(\frac{1}{2}\right)^3 \), so:
\[ \ln\left(\frac{1}{2}\right)^3 = -kt \]
\[ 3\ln\left(\frac{1}{2}\right) = -kt \]
Substitute the value of \( \ln(1/2) = -20k \) back into the equation:
\[ 3(-20k) = -kt \]
\[ -60k = -kt \]
Dividing both sides by \( -k \):
\[ t = 60 \text{ minutes} \]
This means the total time elapsed from the beginning is 60 minutes.
Step 4: Final Answer:
By applying the exponential decay model of Newton's Law of Cooling and recognizing the logarithmic ratio between successive temperature differences, we calculated that the body takes 60 minutes to reach \( 30^{\circ}\text{C} \). This corresponds to Option (C).