According to Newton's law, rate of cooling is proportional to the difference between the temperature of the body and the temperature of the air. If the temperature of the air is `20^(@)C` and body cools for 20 min from `100^(@)C` to `60^(@)C` then the time it will take for it temperature to drop to `30^(@)` is

Options

A. 30 min
B. 40 min
C. 60 min
D. 80 min

Answer

To solve the problem according to Newton's law of cooling, we will follow these steps:

### Step 1: Set Up the Differential Equation
According to Newton's law of cooling, the rate of change of temperature of the body (dT/dt) is proportional to the difference between the temperature of the body (T_b) and the temperature of the air (T_a). This can be expressed as:

\[ \frac{dT}{dt} = -k (T_b - T_a) \]

Given that the temperature of the air \( T_a = 20^\circ C \), we can rewrite the equation as:

\[ \frac{dT}{dt} = -k (T_b - 20) \]

### Step 2: Separate Variables and Integrate
Rearranging the equation gives:

\[ \frac{dT}{T_b - 20} = -k \, dt \]

Now, we integrate both sides:

\[ \int \frac{dT}{T_b - 20} = -k \int dt \]

The left side integrates to:

\[ \ln |T_b - 20| = -kt + C \]

### Step 3: Solve for the Constant of Integration
To find the constant \( C \), we use the initial condition. When \( t = 0 \), \( T_b = 100^\circ C \):

\[ \ln |100 - 20| = C \implies C = \ln 80 \]

Thus, our equation becomes:

\[ \ln |T_b - 20| = -kt + \ln 80 \]

### Step 4: Simplify the Equation
We can exponentiate both sides to eliminate the logarithm:

\[ |T_b - 20| = 80 e^{-kt} \]

### Step 5: Use the Given Condition
We know that after 20 minutes, the temperature drops to \( 60^\circ C \):

\[ |60 - 20| = 80 e^{-20k} \]

This simplifies to:

\[ 40 = 80 e^{-20k} \implies e^{-20k} = \frac{1}{2} \]

Taking the natural logarithm gives:

\[ -20k = \ln \frac{1}{2} \implies k = -\frac{\ln 2}{20} \]

### Step 6: Find the Time for the Temperature to Drop to \( 30^\circ C \)
Now we need to find the time \( t_{30} \) when \( T_b = 30^\circ C \):

\[ |30 - 20| = 80 e^{-kt_{30}} \]

This simplifies to:

\[ 10 = 80 e^{-kt_{30}} \implies e^{-kt_{30}} = \frac{1}{8} \]

Taking the natural logarithm gives:

\[ -kt_{30} = \ln \frac{1}{8} \implies t_{30} = -\frac{\ln \frac{1}{8}}{k} \]

Substituting \( k = -\frac{\ln 2}{20} \):

\[ t_{30} = -\frac{\ln \frac{1}{8}}{-\frac{\ln 2}{20}} = \frac{20 \ln 8}{\ln 2} \]

Since \( \ln 8 = 3 \ln 2 \):

\[ t_{30} = \frac{20 \cdot 3 \ln 2}{\ln 2} = 60 \text{ minutes} \]

### Final Answer
It will take **60 minutes** for the temperature of the body to drop to \( 30^\circ C \).

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