A body takes 5 minutes for cooling from `50^(@)C ` to `40^(@)C` Its temperature comes down to `33.33^(@)C` in next 5 minutes. Temperature of surroundings is

Options

A. (a)`15^(@)C`
B. (b)`20^(@)C`
C. (c)`25^(@)C`
D. (d)`10^(@)C`

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 (surrounding temperature).

### Step-by-Step Solution:

1. **Identify the Initial and Final Temperatures**:
- In the first 5 minutes, the body cools from \(50^\circ C\) to \(40^\circ C\).
- In the next 5 minutes, the body cools from \(40^\circ C\) to \(33.33^\circ C\).

2. **Calculate the Rate of Change of Temperature for the First Interval**:
- The change in temperature (\(dT\)) is \(50^\circ C - 40^\circ C = 10^\circ C\).
- The time interval (\(dt\)) is \(5\) minutes.
- Therefore, the rate of change of temperature is:
\[ \frac{dT}{dt} = \frac{10^\circ C}{5 \text{ min}} = 2^\circ C/\text{min} \]

3. **Calculate the Average Temperature for the First Interval**:
- The average temperature (\(T_{avg1}\)) during this interval is:
\[ T_{avg1} = \frac{50^\circ C + 40^\circ C}{2} = 45^\circ C \]

4. **Set Up the Equation Using Newton's Law of Cooling**:
- According to Newton's Law:
\[ \frac{dT}{dt} = k(T_{avg} - T_{surroundings}) \]
- For the first interval:
\[ 2 = k(45 - T_{surroundings}) \tag{1} \]

5. **Calculate the Rate of Change of Temperature for the Second Interval**:
- The change in temperature (\(dT\)) is \(40^\circ C - 33.33^\circ C = 6.67^\circ C\).
- The time interval (\(dt\)) is again \(5\) minutes.
- Therefore, the rate of change of temperature is:
\[ \frac{dT}{dt} = \frac{6.67^\circ C}{5 \text{ min}} \approx 1.334^\circ C/\text{min} \]

6. **Calculate the Average Temperature for the Second Interval**:
- The average temperature (\(T_{avg2}\)) during this interval is:
\[ T_{avg2} = \frac{40^\circ C + 33.33^\circ C}{2} \approx 36.665^\circ C \]

7. **Set Up the Second Equation Using Newton's Law**:
- For the second interval:
\[ 1.334 = k(36.665 - T_{surroundings}) \tag{2} \]

8. **Divide Equation (1) by Equation (2)**:
- Dividing the two equations to eliminate \(k\):
\[ \frac{2}{1.334} = \frac{45 - T_{surroundings}}{36.665 - T_{surroundings}} \]

9. **Cross Multiply and Solve for \(T_{surroundings}\)**:
- Cross multiplying gives:
\[ 2(36.665 - T_{surroundings}) = 1.334(45 - T_{surroundings}) \]
- Expanding both sides:
\[ 73.33 - 2T_{surroundings} = 60.03 - 1.334T_{surroundings} \]
- Rearranging gives:
\[ 73.33 - 60.03 = 2T_{surroundings} - 1.334T_{surroundings} \]
\[ 13.3 = 0.666T_{surroundings} \]
- Solving for \(T_{surroundings}\):
\[ T_{surroundings} \approx \frac{13.3}{0.666} \approx 20^\circ C \]

### Final Answer:
The temperature of the surroundings is approximately \(20^\circ C\).