A cup of tea cools from `80^(@)C" to "60^(@)C` in 40 seconds. The ambient temperature is `30^(@)C`. In cooling from `60^(@)C" to "50^(@)C`, it will take time:

Options

A. 35 s
B. 30 s
C. 32 s
D. 48 s

Answer

To solve the problem of how long it takes for a cup of tea to cool from 60°C to 50°C, we can use Newton's Law of Cooling. Here's the step-by-step solution:

### Step 1: Understand the Problem
We know that the cup of tea cools from 80°C to 60°C in 40 seconds, and the ambient temperature is 30°C. We need to find out how long it will take to cool from 60°C to 50°C.

### Step 2: Apply Newton's Law of Cooling
According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature.

### Step 3: Calculate the Average Temperature
For the first cooling interval (from 80°C to 60°C):
- Initial temperature (T1) = 80°C
- Final temperature (T2) = 60°C
- Average temperature (T_avg1) = (T1 + T2) / 2 = (80 + 60) / 2 = 70°C

For the second cooling interval (from 60°C to 50°C):
- Initial temperature (T3) = 60°C
- Final temperature (T4) = 50°C
- Average temperature (T_avg2) = (T3 + T4) / 2 = (60 + 50) / 2 = 55°C

### Step 4: Determine Temperature Differences
Now, we calculate the temperature differences:
- For the first interval (80°C to 60°C):
- ΔT1 = T1 - Ambient Temperature = 80°C - 30°C = 50°C
- ΔT2 = T2 - Ambient Temperature = 60°C - 30°C = 30°C

- For the second interval (60°C to 50°C):
- ΔT3 = T3 - Ambient Temperature = 60°C - 30°C = 30°C
- ΔT4 = T4 - Ambient Temperature = 50°C - 30°C = 20°C

### Step 5: Set Up the Proportionality
Using the cooling law, we can set up the following equations based on the time taken:
- For the first cooling interval:
\[ \frac{ΔT1}{ΔT2} = \frac{t1}{t2} \]
where \( t1 = 40 \) seconds (time taken to cool from 80°C to 60°C) and \( t2 \) is the time taken to cool from 60°C to 50°C.

### Step 6: Substitute Values
Substituting the values we have:
\[ \frac{50}{30} = \frac{40}{t2} \]

### Step 7: Solve for \( t2 \)
Cross-multiplying gives:
\[ 50 \cdot t2 = 30 \cdot 40 \]
\[ 50 \cdot t2 = 1200 \]
\[ t2 = \frac{1200}{50} = 24 \text{ seconds} \]

### Step 8: Conclusion
Thus, the time taken for the tea to cool from 60°C to 50°C is 24 seconds.