A body cools in 10 minutes from `60^@C` to `40^@C`. What will be its temperature after next 10 minutes? The temperature of the surroundings is `10^@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 own temperature and the ambient temperature.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial temperature of the body, \( T_1 = 60^\circ C \)
- Final temperature after 10 minutes, \( T_2 = 40^\circ C \)
- Surrounding temperature, \( T_a = 10^\circ C \)
- Time interval, \( t = 10 \) minutes

2. **Calculate the Cooling Constant (α):**
Using the formula from Newton's Law of Cooling:
\[ \frac{T_1 - T_2}{t} = \alpha \left( \frac{T_1 + T_2}{2} - T_a \right) \]
Substituting the known values:
\[ \frac{60 - 40}{10} = \alpha \left( \frac{60 + 40}{2} - 10 \right) \]
Simplifying:
\[ \frac{20}{10} = \alpha \left( 50 - 10 \right) \]
\[ 2 = \alpha \cdot 40 \]
Therefore:
\[ \alpha = \frac{2}{40} = \frac{1}{20} \text{ min}^{-1} \]

3. **Calculate the Temperature After the Next 10 Minutes:**
Now we need to find the temperature \( T_3 \) after another 10 minutes (total of 20 minutes from the start). We will use the same formula:
\[ \frac{T_2 - T_3}{10} = \alpha \left( \frac{T_2 + T_3}{2} - T_a \right) \]
Substituting \( T_2 = 40^\circ C \):
\[ \frac{40 - T_3}{10} = \frac{1}{20} \left( \frac{40 + T_3}{2} - 10 \right) \]
Multiplying both sides by 20:
\[ 2(40 - T_3) = \frac{40 + T_3}{2} - 10 \]
Simplifying:
\[ 80 - 2T_3 = \frac{40 + T_3}{2} - 10 \]
Multiplying through by 2 to eliminate the fraction:
\[ 160 - 4T_3 = 40 + T_3 - 20 \]
Rearranging gives:
\[ 160 - 20 - 40 = 5T_3 \]
\[ 100 = 5T_3 \]
Therefore:
\[ T_3 = 20^\circ C \]

4. **Final Answer:**
The temperature of the body after the next 10 minutes will be \( 20^\circ C \).