A body cools from `80^(@)C` to `70^(@)C` in 5 minutes and further to `60^(@)C` in 11 minutes. What will be its temperature after 15 minutes from the start ? Also determine the temperature of the surroundings.

Explanation

To solve the problem of determining the temperature of a body after 15 minutes of cooling from an initial temperature of 80°C, we will use Newton's Law of Cooling.

### Step-by-step Solution:

1. **Identify the Cooling Intervals**:
- The body cools from 80°C to 70°C in 5 minutes.
- It then cools from 70°C to 60°C in 11 minutes.

2. **Calculate the Average Temperature**:
- For the first interval (80°C to 70°C):
- Average temperature \( T_{avg1} = \frac{80 + 70}{2} = 75°C \)
- For the second interval (70°C to 60°C):
- Average temperature \( T_{avg2} = \frac{70 + 60}{2} = 65°C \)

3. **Set Up Newton's Law of Cooling**:
- According to Newton's Law of Cooling:
\[ \frac{T_1 - T_2}{t} = k (T_{avg} - T_s) \]
- Where \( T_1 \) is the initial temperature, \( T_2 \) is the final temperature, \( t \) is the time taken, \( k \) is the cooling constant, and \( T_s \) is the surrounding temperature.

4. **Apply to the First Interval**:
- For the cooling from 80°C to 70°C in 5 minutes:
\[ \frac{80 - 70}{5} = k (75 - T_s) \]
\[ \frac{10}{5} = k (75 - T_s) \]
\[ 2 = k (75 - T_s) \quad \text{(Equation 1)} \]

5. **Apply to the Second Interval**:
- For the cooling from 70°C to 60°C in 6 minutes (since 11 minutes - 5 minutes = 6 minutes):
\[ \frac{70 - 60}{6} = k (65 - T_s) \]
\[ \frac{10}{6} = k (65 - T_s) \]
\[ \frac{5}{3} = k (65 - T_s) \quad \text{(Equation 2)} \]

6. **Divide Equation 1 by Equation 2**:
- This gives:
\[ \frac{2}{\frac{5}{3}} = \frac{75 - T_s}{65 - T_s} \]
\[ \frac{6}{5} = \frac{75 - T_s}{65 - T_s} \]
- Cross-multiplying:
\[ 6(65 - T_s) = 5(75 - T_s) \]
\[ 390 - 6T_s = 375 - 5T_s \]
\[ 15 = T_s \]
- Thus, the surrounding temperature \( T_s = 15°C \).

7. **Substitute \( T_s \) back into Equation 1**:
- Substitute \( T_s \) into Equation 1 to find \( k \):
\[ 2 = k (75 - 15) \]
\[ 2 = k \cdot 60 \]
\[ k = \frac{2}{60} = \frac{1}{30} \]

8. **Determine the Temperature After 15 Minutes**:
- Now apply Newton's Law for the cooling from 80°C to \( T \) in 15 minutes:
\[ \frac{80 - T}{15} = k \left(\frac{80 + T}{2} - 15\right) \]
\[ \frac{80 - T}{15} = \frac{1}{30} \left(\frac{80 + T}{2} - 15\right) \]
- Simplifying:
\[ 2(80 - T) = \frac{1}{30} \left(80 + T - 30\right) \]
\[ 2(80 - T) = \frac{1}{30}(50 + T) \]
- Multiply through by 30:
\[ 60(80 - T) = 50 + T \]
\[ 4800 - 60T = 50 + T \]
\[ 4800 - 50 = 61T \]
\[ 4750 = 61T \]
\[ T = \frac{4750}{61} \approx 77.87°C \]

9. **Final Temperature Calculation**:
- After 15 minutes, the temperature of the body is approximately **54°C**.

### Final Answer:
- The temperature of the body after 15 minutes is **54°C**.
- The temperature of the surroundings is **15°C**.