Water at ` 100^(@) C` cools in 10 minutes to ` 88^(@)C` in a room temperature of `25^(@)C` .Find the temperature of water after 20 minutes.

Answer

To solve the problem of finding the temperature of water after 20 minutes using Newton's Law of Cooling, we can follow these steps:

### Step 1: Set up the differential equation
According to Newton's Law of Cooling, the rate of change of temperature is proportional to the difference between the temperature of the object and the ambient temperature. Mathematically, this can be expressed as:

\[ \frac{dT}{dt} = -k(T - T_0) \]

where:
- \( T \) is the temperature of the object (water in this case),
- \( T_0 \) is the ambient temperature (room temperature),
- \( k \) is a positive constant.

In this problem, we have:
- Initial temperature \( T(0) = 100^\circ C \)
- Room temperature \( T_0 = 25^\circ C \)

### Step 2: Rewrite the equation
Rearranging the equation gives us:

\[ \frac{dT}{T - T_0} = -k \, dt \]

Substituting \( T_0 = 25 \):

\[ \frac{dT}{T - 25} = -k \, dt \]

### Step 3: Integrate both sides
Integrating both sides, we get:

\[ \int \frac{dT}{T - 25} = -k \int dt \]

This results in:

\[ \ln |T - 25| = -kt + C \]

where \( C \) is the constant of integration.

### Step 4: Solve for the constant of integration
To find \( C \), we use the initial condition \( T(0) = 100 \):

\[ \ln |100 - 25| = C \implies \ln 75 = C \]

Thus, the equation becomes:

\[ \ln |T - 25| = -kt + \ln 75 \]

### Step 5: Solve for \( T \)
Exponentiating both sides gives:

\[ |T - 25| = 75 e^{-kt} \]

Since \( T > 25 \), we can drop the absolute value:

\[ T - 25 = 75 e^{-kt} \implies T = 75 e^{-kt} + 25 \]

### Step 6: Find \( k \) using the second condition
We know that after 10 minutes, the temperature is \( T(10) = 88^\circ C \):

\[ 88 = 75 e^{-10k} + 25 \]

Subtracting 25 from both sides:

\[ 63 = 75 e^{-10k} \]

Dividing both sides by 75:

\[ e^{-10k} = \frac{63}{75} = 0.84 \]

Taking the natural logarithm:

\[ -10k = \ln(0.84) \implies k = -\frac{\ln(0.84)}{10} \]

Calculating \( k \):

\[ k \approx \frac{0.174}{10} \approx 0.0174 \]

### Step 7: Find the temperature after 20 minutes
Now, substituting \( t = 20 \) into the equation:

\[ T = 75 e^{-20k} + 25 \]

Substituting \( k \):

\[ T = 75 e^{-20 \times 0.0174} + 25 \]

Calculating \( e^{-20 \times 0.0174} \):

\[ e^{-0.348} \approx 0.706 \]

Thus,

\[ T = 75 \times 0.706 + 25 \approx 52.95 + 25 \approx 77.95 \]

### Final Answer
The temperature of the water after 20 minutes is approximately:

\[ \boxed{77.95^\circ C} \]