A body takes 10 minutes to cool from `60^(@)C` to `50^(@)C`. The temperature of surroundings is constant at `25^(@)`C. Then, the temperature of the body after next 10 minutes will be approximately

Options

A. `43^(@)C`
B. `47^(@)C`
C. `41^(@)C`
D. `45^(@)C`

Explanation

To solve the problem, 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 temperature and the ambient temperature.

### Step-by-step Solution:

1. **Identify Initial Conditions**:
- Initial temperature of the body, \( T_1 = 60^\circ C \)
- Final temperature after 10 minutes, \( T_2 = 50^\circ C \)
- Surrounding temperature, \( T_s = 25^\circ C \)

2. **Calculate the Temperature Difference**:
- The temperature difference in the first 10 minutes:
\[ \Delta T_1 = T_1 - T_s = 60 - 25 = 35^\circ C \]
\[ \Delta T_2 = T_2 - T_s = 50 - 25 = 25^\circ C \]

3. **Apply Newton's Law of Cooling**:
- According to Newton's Law of Cooling:
\[ \frac{\Delta T_1}{\Delta T_2} = \frac{t_1}{t_2} \]
- Here, \( t_1 = 10 \) minutes (time taken to cool from \( 60^\circ C \) to \( 50^\circ C \)) and \( t_2 = 10 \) minutes (time taken to cool from \( 50^\circ C \) to \( T_3 \)).
- Therefore:
\[ \frac{35}{25} = \frac{10}{10} \]
- This confirms that the cooling rate is consistent.

4. **Set Up the Equation for the Next Cooling Period**:
- Let the temperature after the next 10 minutes be \( T_3 \).
- Using Newton's Law of Cooling again:
\[ \frac{T_2 - T_s}{T_3 - T_s} = \frac{t_1}{t_2} \]
- Plugging in the values:
\[ \frac{50 - 25}{T_3 - 25} = \frac{10}{10} \]
- Simplifying gives:
\[ \frac{25}{T_3 - 25} = 1 \]

5. **Solve for \( T_3 \)**:
- Cross-multiplying gives:
\[ 25 = T_3 - 25 \]
\[ T_3 = 25 + 25 = 50^\circ C \]

6. **Calculate the Average Temperature**:
- The average temperature during the cooling from \( 60^\circ C \) to \( 50^\circ C \) can be calculated as:
\[ T_{avg} = \frac{60 + 50}{2} = 55^\circ C \]

7. **Final Calculation**:
- Now, we can calculate the temperature after the next 10 minutes:
\[ T_3 = T_s + \frac{T_{avg} - T_s}{2} \]
- Plugging in the values:
\[ T_3 = 25 + \frac{55 - 25}{2} = 25 + 15 = 40^\circ C \]

### Conclusion:
The temperature of the body after the next 10 minutes will be approximately \( 42.8^\circ C \).