If a body cools down from `80^(@) C`to `60^(@) C` in 10 min when the temperature of the surrounding of the is `30^(@) C` . Then, the temperature of the body after next 10 min will be

Options

A. `50^(@) C`
B. `48^(@) C`
C. `30^(@) C`
D. None of the above

Explanation

To solve the problem, we will use Newton's Law of Cooling. According to this law, the rate of change of temperature of a body is proportional to the difference between its temperature and the temperature of the surroundings.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial temperature of the body, \( \theta_1 = 80^\circ C \)
- Final temperature of the body after 10 minutes, \( \theta_2 = 60^\circ C \)
- Surrounding temperature, \( \theta_s = 30^\circ C \)
- Time interval, \( t = 10 \) minutes

2. **Apply Newton's Law of Cooling:**
The formula for Newton's Law of Cooling is given by:
\[ \frac{\theta_1 - \theta_2}{t} = \alpha \left( \frac{\theta_1 + \theta_2}{2} - \theta_s \right) \]
Where \( \alpha \) is the cooling constant.

3. **Calculate the left side of the equation:**
\[ \theta_1 - \theta_2 = 80 - 60 = 20^\circ C \]
Therefore,
\[ \frac{20}{10} = 2 \]

4. **Calculate the average temperature:**
\[ \frac{\theta_1 + \theta_2}{2} = \frac{80 + 60}{2} = \frac{140}{2} = 70^\circ C \]

5. **Substitute the values into the equation:**
\[ 2 = \alpha \left( 70 - 30 \right) \]
Simplifying gives:
\[ 2 = \alpha \cdot 40 \]

6. **Solve for \( \alpha \):**
\[ \alpha = \frac{2}{40} = \frac{1}{20} \text{ min}^{-1} \]

7. **Now, calculate the temperature after the next 10 minutes:**
We now consider the new initial temperature \( \theta_2 = 60^\circ C \) and let the final temperature after the next 10 minutes be \( \theta \).

Using the same formula:
\[ \frac{60 - \theta}{10} = \alpha \left( \frac{60 + \theta}{2} - 30 \right) \]

8. **Substituting \( \alpha \) into the equation:**
\[ \frac{60 - \theta}{10} = \frac{1}{20} \left( \frac{60 + \theta}{2} - 30 \right) \]

9. **Simplifying the right side:**
\[ \frac{60 - \theta}{10} = \frac{1}{20} \left( \frac{60 + \theta - 60}{2} \right) = \frac{1}{20} \left( \frac{\theta}{2} \right) = \frac{\theta}{40} \]

10. **Cross-multiply to solve for \( \theta \):**
\[ 40(60 - \theta) = 10\theta \]
\[ 2400 - 40\theta = 10\theta \]
\[ 2400 = 50\theta \]
\[ \theta = \frac{2400}{50} = 48^\circ C \]

### Final Answer:
The temperature of the body after the next 10 minutes will be \( 48^\circ C \).