A body cools from `50.0^(@)C` to `48^(@)C` in 5s. How long will it take to cool from `40.0^(@)C` to `39^(@)C`? Assume the temperature of surroundings to be `30.0^(@)C` and Newton's law of cooling to be valid.

Options

A. 2.5s
B. 10s
C. 20s
D. 5s

Explanation

To solve the problem, we will apply 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 the Given Data:**
- Initial temperature (T1) = 50.0°C
- Final temperature (T2) = 48.0°C
- Time taken (t1) = 5 seconds
- Surrounding temperature (T_s) = 30.0°C

2. **Calculate the Temperature Difference:**
- The temperature difference when cooling from 50°C to 48°C:
\[ \Delta T_1 = T1 - T_s = 50.0°C - 30.0°C = 20.0°C \]
- The average temperature during this cooling:
\[ T_{avg1} = \frac{T1 + T2}{2} = \frac{50.0°C + 48.0°C}{2} = 49.0°C \]

3. **Using Newton's Law of Cooling:**
- According to Newton's Law of Cooling:
\[ \frac{dT}{dt} = -k(T - T_s) \]
- Rearranging gives:
\[ \Delta T = k \cdot (T_{avg1} - T_s) \cdot t1 \]
- For the first cooling process:
\[ 2°C = k \cdot (49.0°C - 30.0°C) \cdot 5s \]
\[ 2°C = k \cdot 19.0°C \cdot 5s \]
\[ k = \frac{2°C}{95°C \cdot s} = \frac{2}{95} \text{ s}^{-1} \]

4. **Cooling from 40.0°C to 39.0°C:**
- Now, we need to find the time taken (t2) to cool from 40.0°C to 39.0°C.
- The temperature difference:
\[ \Delta T_2 = T3 - T_s = 40.0°C - 30.0°C = 10.0°C \]
- The average temperature during this cooling:
\[ T_{avg2} = \frac{40.0°C + 39.0°C}{2} = 39.5°C \]
- Using Newton's Law of Cooling again:
\[ 1°C = k \cdot (39.5°C - 30.0°C) \cdot t2 \]
\[ 1°C = k \cdot 9.5°C \cdot t2 \]
- Substituting the value of k:
\[ 1°C = \left(\frac{2}{95}\right) \cdot 9.5°C \cdot t2 \]
\[ t2 = \frac{1°C \cdot 95}{2 \cdot 9.5} \]
\[ t2 = \frac{95}{19} = 5 \text{ seconds} \]

### Final Answer:
It will take **5 seconds** to cool from 40.0°C to 39.0°C.