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 temperature and the ambient temperature.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature of the body, \( T_1 = 60^\circ C \)
- Final temperature after 10 minutes, \( T_2 = 40^\circ C \)
- Surrounding temperature, \( T_0 = 10^\circ C \)
- Time interval, \( t = 10 \) minutes

2. **Apply Newton's Law of Cooling:**
The formula according to Newton's Law of Cooling is:
\[ \frac{T_1 - T_2}{t} = \alpha \left( \frac{T_1 + T_2}{2} - T_0 \right) \]
Plugging in the values:
\[ \frac{60 - 40}{10} = \alpha \left( \frac{60 + 40}{2} - 10 \right) \]
Simplifying the left side:
\[ \frac{20}{10} = \alpha \left( 50 - 10 \right) \]
This simplifies to:
\[ 2 = \alpha \cdot 40 \]

3. **Solve for \( \alpha \):**
Rearranging gives:
\[ \alpha = \frac{2}{40} = \frac{1}{20} \]

4. **Find the Temperature After the Next 10 Minutes:**
Now we need to find the temperature \( T_3 \) after the next 10 minutes, where \( T_2 = 40^\circ C \) is the new initial temperature.
Using the same formula:
\[ \frac{T_2 - T_3}{10} = \alpha \left( \frac{T_2 + T_3}{2} - T_0 \right) \]
Plugging in the values:
\[ \frac{40 - T_3}{10} = \frac{1}{20} \left( \frac{40 + T_3}{2} - 10 \right) \]
Simplifying the left side:
\[ 4 - \frac{T_3}{10} = \frac{1}{20} \left( 20 + \frac{T_3}{2} - 10 \right) \]
This simplifies to:
\[ 4 - \frac{T_3}{10} = \frac{1}{20} \left( 10 + \frac{T_3}{2} \right) \]
Multiplying both sides by 20 to eliminate the fraction:
\[ 80 - 2T_3 = 10 + \frac{T_3}{2} \]

5. **Rearranging the Equation:**
Bringing all \( T_3 \) terms to one side:
\[ 80 - 10 = 2T_3 + \frac{T_3}{2} \]
This simplifies to:
\[ 70 = \frac{4T_3 + T_3}{2} \]
Which leads to:
\[ 70 = \frac{5T_3}{2} \]

6. **Solving for \( T_3 \):**
Multiplying both sides by 2:
\[ 140 = 5T_3 \]
Dividing by 5 gives:
\[ T_3 = 28^\circ C \]

### Final Answer:
The temperature of the body after the next 10 minutes will be \( T_3 = 28^\circ C \).