A liquid cools from `50^(@)C` to `45^(@)C` in 5 minutes and from `45^(@)C` to `41.5^(@)C` in the next 5 minutes. The temperature of the surrounding is

Options

A. `27^(@)C`
B. `40.3^(@)C`
C. `23.3^(@)C`
D. `33.3^(@)C`

Explanation

To solve the problem of determining the temperature of the surroundings (θ₀) using Newton's law of cooling, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Cooling Process**:
The liquid cools from 50°C to 45°C in 5 minutes, and then from 45°C to 41.5°C in the next 5 minutes. We will apply Newton's law of cooling for both intervals.

2. **Applying Newton's Law of Cooling**:
According to Newton's law of cooling:
\[ \frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - \theta_0 \right) \]
where \(T_1\) is the initial temperature, \(T_2\) is the final temperature, \(t\) is the time, \(k\) is the cooling constant, and \(\theta_0\) is the surrounding temperature.

3. **First Cooling Interval (50°C to 45°C)**:
- Here, \(T_1 = 50°C\), \(T_2 = 45°C\), and \(t = 5\) minutes.
\[ \frac{50 - 45}{5} = k \left( \frac{50 + 45}{2} - \theta_0 \right) \]
Simplifying:
\[ 1 = k \left( 47.5 - \theta_0 \right) \]
Rearranging gives us:
\[ 2 = k (95 - 2\theta_0) \quad \text{(Equation 1)} \]

4. **Second Cooling Interval (45°C to 41.5°C)**:
- Here, \(T_1 = 45°C\), \(T_2 = 41.5°C\), and \(t = 5\) minutes.
\[ \frac{45 - 41.5}{5} = k \left( \frac{45 + 41.5}{2} - \theta_0 \right) \]
Simplifying:
\[ 0.5 = k \left( 43.25 - \theta_0 \right) \]
Rearranging gives us:
\[ \frac{1}{10} = k (86.5 - 2\theta_0) \quad \text{(Equation 2)} \]

5. **Dividing the Two Equations**:
We can divide Equation 1 by Equation 2:
\[ \frac{2}{\frac{1}{10}} = \frac{k(95 - 2\theta_0)}{k(86.5 - 2\theta_0)} \]
Simplifying gives:
\[ 20 = \frac{95 - 2\theta_0}{86.5 - 2\theta_0} \]

6. **Cross Multiplying**:
\[ 20(86.5 - 2\theta_0) = 95 - 2\theta_0 \]
Expanding:
\[ 1730 - 40\theta_0 = 95 - 2\theta_0 \]

7. **Rearranging Terms**:
Bringing all terms involving \(\theta_0\) to one side:
\[ 40\theta_0 - 2\theta_0 = 1730 - 95 \]
Simplifying gives:
\[ 38\theta_0 = 1635 \]

8. **Solving for \(\theta_0\)**:
\[ \theta_0 = \frac{1635}{38} \approx 43.0°C \]

9. **Final Answer**:
The temperature of the surroundings is approximately \(43.0°C\).