The temperature of a liquid drops from 365 K to 361 K in 2 minutes . Find the time during which temperature of the liquid drops from 344 K to 342 K.Temperature of room is 293 K

Options

A. 84 sec
B. 72 sec
C. 66 sec
D. 60 sec

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 own temperature and the ambient temperature.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Initial temperature of the liquid (T1) = 365 K
- Final temperature of the liquid (T2) = 361 K
- Time taken for this change (t1) = 2 minutes
- Ambient temperature (Ta) = 293 K

2. **Calculate the Temperature Difference:**
- The average temperature of the liquid during the first observation:
\[ T_{\text{avg1}} = \frac{T1 + T2}{2} = \frac{365 + 361}{2} = 363 K \]
- The temperature difference between the average temperature and the ambient temperature:
\[ \Delta T_1 = T_{\text{avg1}} - Ta = 363 - 293 = 70 K \]

3. **Calculate the Rate of Cooling (K):**
- Using Newton's Law of Cooling:
\[ \frac{T1 - T2}{t1} = K \cdot (T_{\text{avg1}} - Ta) \]
- Substitute the values:
\[ \frac{365 - 361}{2} = K \cdot 70 \]
\[ \frac{4}{2} = K \cdot 70 \implies 2 = K \cdot 70 \implies K = \frac{2}{70} = \frac{1}{35} \text{ min}^{-1} \]

4. **Determine the Time for the Second Temperature Drop:**
- For the second observation:
- Initial temperature (T3) = 344 K
- Final temperature (T4) = 342 K
- Calculate the average temperature:
\[ T_{\text{avg2}} = \frac{T3 + T4}{2} = \frac{344 + 342}{2} = 343 K \]
- Calculate the temperature difference:
\[ \Delta T_2 = T_{\text{avg2}} - Ta = 343 - 293 = 50 K \]

5. **Use K to Find Time (t2) for the Second Drop:**
- Using Newton's Law of Cooling again:
\[ \frac{T3 - T4}{t2} = K \cdot (T_{\text{avg2}} - Ta) \]
- Substitute the values:
\[ \frac{344 - 342}{t2} = \frac{1}{35} \cdot 50 \]
\[ \frac{2}{t2} = \frac{50}{35} \implies t2 = \frac{2 \cdot 35}{50} = \frac{70}{50} = 1.4 \text{ minutes} \]

### Final Answer:
The time during which the temperature of the liquid drops from 344 K to 342 K is **1.4 minutes**.