A body takes 4 minutes to cool from `100^(@)C` to `70^(@)C`. To cool from `70^(@)C` to `40^(@)C` it will take (room temperture os `15^(@)C`)

Options

A. 7 minutes
B. 6 minutes
C. 5 minutes
D. 4 minutes

Answer

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 (room temperature).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial temperature (T1) = 100°C
- Final temperature after 4 minutes (T2) = 70°C
- Room temperature (T_room) = 15°C
- Time taken to cool from T1 to T2 (t1) = 4 minutes

2. **Apply Newton's Law of Cooling:**
The formula according to Newton's Law of Cooling is:
\[ \frac{T_1 - T_2}{t} = k \cdot (T_{avg} - T_{room}) \]
Where:
- \(T_{avg} = \frac{T_1 + T_2}{2}\)

3. **Calculate the Average Temperature (T_avg):**
\[ T_{avg} = \frac{100 + 70}{2} = \frac{170}{2} = 85°C \]

4. **Substitute the Values into the Formula:**
\[ \frac{100 - 70}{4} = k \cdot (85 - 15) \]
\[ \frac{30}{4} = k \cdot 70 \]
\[ 7.5 = k \cdot 70 \]
\[ k = \frac{7.5}{70} = \frac{3}{28} \]

5. **Now, Calculate the Time to Cool from 70°C to 40°C:**
- Initial temperature (T3) = 70°C
- Final temperature (T4) = 40°C
- Average temperature for this interval:
\[ T_{avg2} = \frac{70 + 40}{2} = \frac{110}{2} = 55°C \]

6. **Set Up the Equation Using Newton's Law of Cooling Again:**
\[ \frac{T_3 - T_4}{t_2} = k \cdot (T_{avg2} - T_{room}) \]
\[ \frac{70 - 40}{t_2} = \frac{3}{28} \cdot (55 - 15) \]
\[ \frac{30}{t_2} = \frac{3}{28} \cdot 40 \]

7. **Calculate the Right Side:**
\[ \frac{3 \cdot 40}{28} = \frac{120}{28} = \frac{30}{7} \]

8. **Set Up the Equation:**
\[ \frac{30}{t_2} = \frac{30}{7} \]

9. **Cross Multiply to Solve for t_2:**
\[ 30 \cdot 7 = 30 \cdot t_2 \]
\[ t_2 = 7 \text{ minutes} \]

### Final Answer:
The time taken to cool from 70°C to 40°C is **7 minutes**.