An object is cooled from `75^(@)C` to `65^(@)C` in 2 min in a room at `30^(@)C`. The time taken to cool the same object from `55^(@)C` to `45^(@)C` in the same room is

Options

A. A) 5 min
B. B) 3 min
C. C) 4 min
D. D) 2 min

Answer

To solve the problem of how long it takes for an object to cool from \(55^\circ C\) to \(45^\circ C\) in a room at \(30^\circ C\), we can use Newton's Law of Cooling. According to this law, the rate of cooling of an object is proportional to the difference in temperature between the object and its surroundings.

### Step-by-Step Solution:

1. **Understanding the Cooling Process**:
- The cooling process can be described by the formula:
\[ \frac{dT}{dt} = -k(T - T_0) \]
where:
- \(T\) is the temperature of the object,
- \(T_0\) is the ambient temperature (room temperature),
- \(k\) is the cooling constant,
- \(dT\) is the change in temperature,
- \(dt\) is the change in time.

2. **Cooling from \(75^\circ C\) to \(65^\circ C\)**:
- Initial temperature \(T_1 = 75^\circ C\)
- Final temperature \(T_2 = 65^\circ C\)
- Room temperature \(T_0 = 30^\circ C\)
- The change in temperature \(\Delta T_1 = T_1 - T_0 = 75 - 30 = 45^\circ C\)
- The change in temperature \(\Delta T_2 = T_2 - T_0 = 65 - 30 = 35^\circ C\)
- The time taken for this cooling is \(t_1 = 2\) minutes.

3. **Using the Cooling Law**:
- From the cooling law, we can set up the ratio of the time taken for the two cooling processes:
\[ \frac{t_1}{t_2} = \frac{\Delta T_1}{\Delta T_2} \]
- Here, \(t_2\) is the time taken to cool from \(55^\circ C\) to \(45^\circ C\).

4. **Calculating \(\Delta T_2\)**:
- For the second cooling process:
- Initial temperature \(T_3 = 55^\circ C\)
- Final temperature \(T_4 = 45^\circ C\)
- The change in temperature \(\Delta T_3 = T_3 - T_0 = 55 - 30 = 25^\circ C\)
- The change in temperature \(\Delta T_4 = T_4 - T_0 = 45 - 30 = 15^\circ C\)

5. **Setting Up the Ratio**:
- Now we can substitute the values into the ratio:
\[ \frac{2}{t_2} = \frac{35}{15} \]

6. **Solving for \(t_2\)**:
- Cross-multiplying gives:
\[ 2 \cdot 15 = 35 \cdot t_2 \]
\[ 30 = 35t_2 \]
\[ t_2 = \frac{30}{35} = \frac{6}{7} \text{ minutes} \]

7. **Final Result**:
- To convert this into a more understandable format, we can approximate:
\[ t_2 \approx 0.857 \text{ minutes} \text{ or } 51.4 \text{ seconds} \]

### Conclusion:
The time taken to cool the object from \(55^\circ C\) to \(45^\circ C\) in a room at \(30^\circ C\) is approximately \(0.857\) minutes or \(51.4\) seconds.