A body cools down from `50^(@)C` to `45^(@)C` in 5 minutes and to `40^(@)C` in another 8 minutes. Find the temperature of the surrounding.

Explanation

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

### Step-by-Step Solution:

1. **Identify Initial and Final Temperatures**:
- The body cools from \( T_1 = 50^\circ C \) to \( T_2 = 45^\circ C \) in 5 minutes.
- Then it cools from \( T_2 = 45^\circ C \) to \( T_3 = 40^\circ C \) in another 8 minutes.

2. **Calculate Average Temperatures**:
- For the first interval (from \( 50^\circ C \) to \( 45^\circ C \)):
\[ T_{avg1} = \frac{T_1 + T_2}{2} = \frac{50 + 45}{2} = 47.5^\circ C \]
- For the second interval (from \( 45^\circ C \) to \( 40^\circ C \)):
\[ T_{avg2} = \frac{T_2 + T_3}{2} = \frac{45 + 40}{2} = 42.5^\circ C \]

3. **Set Up Newton's Law of Cooling**:
- According to Newton's Law of Cooling:
\[ \frac{dT}{dt} = -k(T - T_s) \]
- Where \( T_s \) is the surrounding temperature and \( k \) is the cooling constant.

4. **Calculate Rate of Cooling for Each Interval**:
- For the first interval:
\[ \text{Rate of cooling} = \frac{T_1 - T_2}{\text{time}} = \frac{50 - 45}{5} = 1^\circ C/\text{min} \]
- For the second interval:
\[ \text{Rate of cooling} = \frac{T_2 - T_3}{\text{time}} = \frac{45 - 40}{8} = 0.625^\circ C/\text{min} \]

5. **Set Up Equations Using Average Temperatures**:
- For the first interval:
\[ 1 = k(47.5 - T_s) \quad \text{(Equation 1)} \]
- For the second interval:
\[ 0.625 = k(42.5 - T_s) \quad \text{(Equation 2)} \]

6. **Express \( k \) from Equation 1**:
- Rearranging Equation 1 gives:
\[ k = \frac{1}{47.5 - T_s} \]

7. **Substitute \( k \) into Equation 2**:
- Substitute \( k \) from Equation 1 into Equation 2:
\[ 0.625 = \frac{1}{47.5 - T_s}(42.5 - T_s) \]
- Rearranging this gives:
\[ 0.625(47.5 - T_s) = 42.5 - T_s \]

8. **Solve for \( T_s \)**:
- Distributing \( 0.625 \):
\[ 29.6875 - 0.625T_s = 42.5 - T_s \]
- Rearranging gives:
\[ 29.6875 + T_s = 42.5 + 0.625T_s \]
- Combine like terms:
\[ 29.6875 - 42.5 = 0.625T_s - T_s \]
\[ -12.8125 = -0.375T_s \]
- Solving for \( T_s \):
\[ T_s = \frac{12.8125}{0.375} \approx 34.1^\circ C \]

### Final Answer:
The temperature of the surrounding is approximately \( 34.1^\circ C \).