A body cools down from `45^(@)C` to `40^(@)C` in 5 minutes and to `35^(@)` in another 8 minutes. Find the temperature of the surrounding.

Answer

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

### Step 1: Understand the Problem
We have a body cooling from 45°C to 40°C in 5 minutes and then from 40°C to 35°C in 8 minutes. We need to find the temperature of the surrounding (let's denote it as \( T_0 \)).

### Step 2: Apply Newton's Law of Cooling
Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. The formula can be expressed as:

\[ \frac{\Delta T}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \]

Where:
- \( \Delta T \) is the change in temperature,
- \( t \) is the time taken,
- \( k \) is a constant,
- \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively.

### Step 3: Set Up the Equations
1. For the first cooling period (from 45°C to 40°C in 5 minutes):
- \( T_1 = 45°C \)
- \( T_2 = 40°C \)
- \( t = 5 \) minutes

The equation becomes:
\[ \frac{45 - 40}{5} = k \left( \frac{45 + 40}{2} - T_0 \right) \]
Simplifying gives:
\[ 1 = k \left( 42.5 - T_0 \right) \quad \text{(Equation 1)} \]

2. For the second cooling period (from 40°C to 35°C in 8 minutes):
- \( T_1 = 40°C \)
- \( T_2 = 35°C \)
- \( t = 8 \) minutes

The equation becomes:
\[ \frac{40 - 35}{8} = k \left( \frac{40 + 35}{2} - T_0 \right) \]
Simplifying gives:
\[ \frac{5}{8} = k \left( 37.5 - T_0 \right) \quad \text{(Equation 2)} \]

### Step 4: Solve for \( k \) from Equation 1
From Equation 1:
\[ 1 = k (42.5 - T_0) \]
Thus,
\[ k = \frac{1}{42.5 - T_0} \quad \text{(Equation 3)} \]

### Step 5: Substitute \( k \) in Equation 2
Substituting \( k \) from Equation 3 into Equation 2:
\[ \frac{5}{8} = \frac{1}{42.5 - T_0} \left( 37.5 - T_0 \right) \]

### Step 6: Cross Multiply and Simplify
Cross multiplying gives:
\[ 5(42.5 - T_0) = 8(37.5 - T_0) \]
Expanding both sides:
\[ 212.5 - 5T_0 = 300 - 8T_0 \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ 3T_0 = 300 - 212.5 \]
\[ 3T_0 = 87.5 \]
\[ T_0 = \frac{87.5}{3} = 29.17°C \]

### Final Answer
The temperature of the surrounding is approximately \( 29.17°C \).

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