A bowl filled with very hot soup cools from ` 98^@C` to `86^@C` in 2 minutes when the room temperature is `22^@C`. How long it will take to cool from `75^@C `to `69^@C` ?

Options

A. 2 minutes
B. 1 minutes
C. 1.4 minutes
D. 0.5 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.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial temperature (T1) = 98°C
- Final temperature (T2) = 86°C
- Time taken (t1) = 2 minutes
- Room temperature (Ts) = 22°C

2. **Calculate the Rate of Cooling (R1):**
\[ R_1 = \frac{T_1 - T_2}{t_1} = \frac{98°C - 86°C}{2 \text{ minutes}} = \frac{12°C}{2 \text{ minutes}} = 6°C/\text{min} \]

3. **Determine the Mean Temperature (T) during the first cooling period:**
\[ T = \frac{T_1 + T_2}{2} = \frac{98°C + 86°C}{2} = \frac{184°C}{2} = 92°C \]

4. **Set up the equation using Newton's Law of Cooling for the first interval:**
\[ R_1 = -k(T - T_s) \Rightarrow 6 = -k(92 - 22) \Rightarrow 6 = -k(70) \]
Rearranging gives:
\[ k = -\frac{6}{70} = -\frac{3}{35} \text{ min}^{-1} \]

5. **Now, consider the second cooling period:**
- Initial temperature (T3) = 75°C
- Final temperature (T4) = 69°C
- Let the time taken for this cooling be \( t_2 \).

6. **Calculate the Rate of Cooling (R2) for the second interval:**
\[ R_2 = \frac{T_3 - T_4}{t_2} = \frac{75°C - 69°C}{t_2} = \frac{6°C}{t_2} \]

7. **Determine the Mean Temperature (T) during the second cooling period:**
\[ T = \frac{T_3 + T_4}{2} = \frac{75°C + 69°C}{2} = \frac{144°C}{2} = 72°C \]

8. **Set up the equation using Newton's Law of Cooling for the second interval:**
\[ R_2 = -k(T - T_s) \Rightarrow \frac{6}{t_2} = -k(72 - 22) \Rightarrow \frac{6}{t_2} = -k(50) \]

9. **Substituting the value of k from step 4:**
\[ \frac{6}{t_2} = -\left(-\frac{3}{35}\right)(50) \Rightarrow \frac{6}{t_2} = \frac{150}{35} \]
Rearranging gives:
\[ t_2 = \frac{6 \times 35}{150} = \frac{210}{150} = 1.4 \text{ minutes} \]

### Final Answer:
The time taken to cool from 75°C to 69°C is **1.4 minutes**.