A body initially at `80^(@)`C cools to `64^(@)`C in 5 minutes and to `52^(@)` C in 10 minutes. What will be the temprature after 15 minutes and what is the temperature of the surroundings ?

Explanation

To solve the problem step by step, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature (temperature of the surroundings).

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial temperature of the body, \( T_1 = 80^\circ C \)
- Temperature after 5 minutes, \( T_2 = 64^\circ C \)
- Temperature after 10 minutes, \( T_3 = 52^\circ C \)

2. **Set Up the Equations Using Newton's Law of Cooling:**
- For the first interval (0 to 5 minutes):
\[ \frac{T_1 - T_2}{5} = k (T_{avg1} - T_s) \]
Where \( T_{avg1} = \frac{T_1 + T_2}{2} = \frac{80 + 64}{2} = 72^\circ C \)
Thus, the equation becomes:
\[ \frac{80 - 64}{5} = k (72 - T_s) \]
Simplifying gives:
\[ \frac{16}{5} = k (72 - T_s) \quad \text{(1)} \]

- For the second interval (5 to 10 minutes):
\[ \frac{T_2 - T_3}{5} = k (T_{avg2} - T_s) \]
Where \( T_{avg2} = \frac{T_2 + T_3}{2} = \frac{64 + 52}{2} = 58^\circ C \)
Thus, the equation becomes:
\[ \frac{64 - 52}{5} = k (58 - T_s) \]
Simplifying gives:
\[ \frac{12}{5} = k (58 - T_s) \quad \text{(2)} \]

3. **Divide Equation (1) by Equation (2):**
\[ \frac{\frac{16}{5}}{\frac{12}{5}} = \frac{k(72 - T_s)}{k(58 - T_s)} \]
This simplifies to:
\[ \frac{16}{12} = \frac{72 - T_s}{58 - T_s} \]
Cross-multiplying gives:
\[ 16(58 - T_s) = 12(72 - T_s) \]
Expanding both sides:
\[ 928 - 16T_s = 864 - 12T_s \]
Rearranging gives:
\[ 928 - 864 = 16T_s - 12T_s \]
Thus:
\[ 64 = 4T_s \implies T_s = 16^\circ C \]

4. **Find the Temperature After 15 Minutes:**
- For the third interval (10 to 15 minutes):
\[ \frac{T_3 - T_f}{5} = k (T_{avg3} - T_s) \]
Where \( T_{avg3} = \frac{T_3 + T_f}{2} \)
Thus, the equation becomes:
\[ \frac{52 - T_f}{5} = k \left(\frac{52 + T_f}{2} - 16\right) \]
- We can use the value of \( k \) from either previous equation. Let's use equation (1) to express \( k \):
\[ k = \frac{16}{5(72 - 16)} = \frac{16}{5 \times 56} = \frac{16}{280} = \frac{1}{17.5} \]
- Substituting \( k \) into the third equation:
\[ \frac{52 - T_f}{5} = \frac{1}{17.5} \left(\frac{52 + T_f}{2} - 16\right) \]
- Solving this equation will yield \( T_f \).

5. **Final Calculation:**
- After substituting and simplifying, we find:
\[ T_f = 48^\circ C \]

### Final Answers:
- The temperature of the surroundings, \( T_s = 16^\circ C \)
- The temperature of the body after 15 minutes, \( T_f = 48^\circ C \)