A body cools from `50^@C` to `40^@C` in 5 mintues in surrounding temperature `20^@C`. Find temperature of body in next 5 mintues.

Options

A. `13.3^@C`
B. `23.3^@C`
C. `43.3^@C`
D. `33.3^@C`

Answer

To solve the problem of finding the temperature of the body after the next 5 minutes using Newton's Law of Cooling, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Problem**: The body cools from 50°C to 40°C in 5 minutes, and the surrounding temperature is 20°C. We need to find the temperature of the body after the next 5 minutes.

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 surrounding temperature. Mathematically, this can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_s) \]
where \( T \) is the temperature of the body, \( T_s \) is the surrounding temperature, and \( k \) is a positive constant.

3. **Set Up the Initial Conditions**:
- Initial temperature \( T_0 = 50°C \)
- Final temperature after 5 minutes \( T_1 = 40°C \)
- Surrounding temperature \( T_s = 20°C \)

4. **Calculate the Cooling Constant \( k \)**:
- From the first 5 minutes, we can use the formula derived from integrating Newton's Law:
\[ \ln\left(\frac{T_1 - T_s}{T_0 - T_s}\right) = -kt \]
Plugging in the values:
\[ \ln\left(\frac{40 - 20}{50 - 20}\right) = -k \cdot 5 \]
This simplifies to:
\[ \ln\left(\frac{20}{30}\right) = -5k \]
\[ \ln\left(\frac{2}{3}\right) = -5k \]
Therefore, we can express \( k \):
\[ k = -\frac{1}{5} \ln\left(\frac{2}{3}\right) \]

5. **Calculate the Temperature After the Next 5 Minutes**:
- Now we need to find the temperature after another 5 minutes, using the new initial temperature \( T_1 = 40°C \):
\[ T_2 = T_s + (T_1 - T_s)e^{-kt} \]
Substituting the values:
\[ T_2 = 20 + (40 - 20)e^{-k \cdot 5} \]
We already found \( k \), so:
\[ T_2 = 20 + 20 \cdot e^{-(-\frac{1}{5} \ln\left(\frac{2}{3}\right)) \cdot 5} \]
Simplifying:
\[ T_2 = 20 + 20 \cdot \left(\frac{2}{3}\right) \]
\[ T_2 = 20 + \frac{40}{3} \]
\[ T_2 = 20 + 13.33 = 33.33°C \]

### Final Answer:
The temperature of the body after the next 5 minutes is approximately **33.33°C**.