An object kept in a large room having air temperature of `25^@C` takes 12 minutes to cool from `80^@C` to `70^@C` . The time taken to cool for the same object from `70^@C` to `60^@C` would be nearly,

Options

A. 15 min
B. 10 min
C. 12 min
D. 20 min

Explanation

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) = 80°C
- Final temperature (T2) = 70°C
- Surrounding temperature (Ts) = 25°C
- Time taken (t1) = 12 minutes

2. **Use Newton's Law of Cooling:**
According to Newton's Law of Cooling:
\[ \frac{T1 - T2}{t1} = K \left( \frac{T1 + T2}{2} - Ts \right) \]

3. **Substitute the Values for the First Cooling Interval:**
- Calculate the left side:
\[ T1 - T2 = 80 - 70 = 10 \]
- Calculate the average temperature:
\[ \frac{T1 + T2}{2} = \frac{80 + 70}{2} = 75 \]
- Calculate the difference from the surrounding temperature:
\[ 75 - 25 = 50 \]
- Now, substituting these values into the equation:
\[ \frac{10}{12} = K \cdot 50 \]

4. **Solve for K:**
\[ K = \frac{10}{12 \cdot 50} = \frac{1}{60} \]

5. **Now, Calculate the Time for the Second Cooling Interval (from 70°C to 60°C):**
- Here, T1 = 70°C and T2 = 60°C.
- Using the same formula:
\[ \frac{T1 - T2}{t2} = K \left( \frac{T1 + T2}{2} - Ts \right) \]
- Calculate the left side:
\[ T1 - T2 = 70 - 60 = 10 \]
- Calculate the average temperature:
\[ \frac{T1 + T2}{2} = \frac{70 + 60}{2} = 65 \]
- Calculate the difference from the surrounding temperature:
\[ 65 - 25 = 40 \]
- Now substituting these values into the equation:
\[ \frac{10}{t2} = \frac{1}{60} \cdot 40 \]

6. **Solve for t2:**
\[ \frac{10}{t2} = \frac{40}{60} \]
\[ \frac{10}{t2} = \frac{2}{3} \]
- Cross-multiplying gives:
\[ 10 \cdot 3 = 2 \cdot t2 \]
\[ 30 = 2t2 \]
\[ t2 = \frac{30}{2} = 15 \text{ minutes} \]

### Final Answer:
The time taken to cool from 70°C to 60°C would be nearly **15 minutes**.