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. `15^(@)C`
B. `20^(@)C`
C. `25^(@)C`
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 (temperature of the surroundings).

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial temperature of the body, \( T_1 = 50^\circ C \)
- Temperature after 5 minutes, \( T_2 = 40^\circ C \)
- Temperature after another 5 minutes, \( T_3 = 33.33^\circ C \)

2. **Define Surrounding Temperature:**
- Let the surrounding temperature be \( T_0 \).

3. **Apply Newton's Law of Cooling:**
- According to Newton's Law of Cooling, the rate of cooling can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_0) \]
- For the first interval (from \( T_1 \) to \( T_2 \)):
\[ \frac{T_2 - T_1}{\Delta t_1} = -k(T_1 - T_0) \]
\[ \frac{40 - 50}{5} = -k(50 - T_0) \]
\[ -2 = -k(50 - T_0) \quad \text{(Equation 1)} \]

4. **Calculate for the Second Interval:**
- For the second interval (from \( T_2 \) to \( T_3 \)):
\[ \frac{T_3 - T_2}{\Delta t_2} = -k(T_2 - T_0) \]
\[ \frac{33.33 - 40}{5} = -k(40 - T_0) \]
\[ -1.334 = -k(40 - T_0) \quad \text{(Equation 2)} \]

5. **Set Up the Equations:**
- From Equation 1:
\[ k(50 - T_0) = 2 \quad \text{(1)} \]
- From Equation 2:
\[ k(40 - T_0) = 1.334 \quad \text{(2)} \]

6. **Divide the Equations:**
- Divide Equation (1) by Equation (2):
\[ \frac{2}{1.334} = \frac{50 - T_0}{40 - T_0} \]
- Let’s calculate \( \frac{2}{1.334} \approx 1.5 \):
\[ 1.5 = \frac{50 - T_0}{40 - T_0} \]

7. **Cross Multiply and Solve for \( T_0 \):**
- Cross multiply:
\[ 1.5(40 - T_0) = 50 - T_0 \]
\[ 60 - 1.5T_0 = 50 - T_0 \]
\[ 60 - 50 = 1.5T_0 - T_0 \]
\[ 10 = 0.5T_0 \]
\[ T_0 = 20^\circ C \]

### Final Answer:
The temperature of the surroundings is \( T_0 = 20^\circ C \).