A body cools in 3 minute from 90°C to 80°C. The temperature reduce to 70°C in next (If temperature of surroundings is 20°C)

Options

A. 2.54 minutes
B. 2 minutes
C. 6 minutes
D. 3.54 minutes

Answer

To solve the problem of how long it takes for a body to cool from 80°C to 70°C given that it cools from 90°C to 80°C in 3 minutes, we will use Newton's Law of Cooling.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The body cools from 90°C to 80°C in 3 minutes.
- We want to find out how long it takes to cool from 80°C to 70°C.
- The surrounding temperature (θs) is given as 20°C.

2. **Applying Newton's Law of Cooling**:
- According to Newton's Law of Cooling, the rate of cooling is proportional to the difference between the temperature of the body and the surrounding temperature.
- The formula can be expressed as:
\[ \frac{\Delta T}{\Delta t} = k (T_{avg} - T_s) \]
where:
- \( \Delta T \) is the change in temperature,
- \( \Delta t \) is the time taken,
- \( k \) is the cooling constant,
- \( T_{avg} \) is the average temperature during the cooling period,
- \( T_s \) is the surrounding temperature.

3. **Calculating for the First Cooling Interval (90°C to 80°C)**:
- Here, \( T_1 = 90°C \) and \( T_2 = 80°C \).
- The average temperature \( T_{avg1} = \frac{90 + 80}{2} = 85°C \).
- The change in temperature \( \Delta T_1 = 90 - 80 = 10°C \).
- The time taken \( \Delta t_1 = 3 \) minutes.
- Plugging into the formula:
\[ \frac{10}{3} = k (85 - 20) \]
\[ \frac{10}{3} = k \cdot 65 \]
\[ k = \frac{10}{3 \cdot 65} = \frac{10}{195} \approx 0.0513 \text{ min}^{-1} \]

4. **Calculating for the Second Cooling Interval (80°C to 70°C)**:
- Now, \( T_2 = 80°C \) and \( T_3 = 70°C \).
- The average temperature \( T_{avg2} = \frac{80 + 70}{2} = 75°C \).
- The change in temperature \( \Delta T_2 = 80 - 70 = 10°C \).
- Let the time taken be \( t \).
- Plugging into the formula:
\[ \frac{10}{t} = k (75 - 20) \]
\[ \frac{10}{t} = k \cdot 55 \]
- Substituting the value of \( k \):
\[ \frac{10}{t} = 0.0513 \cdot 55 \]
\[ \frac{10}{t} = 2.8265 \]
\[ t = \frac{10}{2.8265} \approx 3.54 \text{ minutes} \]

5. **Conclusion**:
- The time taken for the body to cool from 80°C to 70°C is approximately **3.54 minutes**.