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 a body cooling from 50°C to 40°C in 5 minutes with a surrounding temperature of 20°C, we will use Newton's Law of Cooling. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understand the Problem:**
- Initial temperature of the body, \( T_1 = 50°C \)
- Final temperature after 5 minutes, \( T_2 = 40°C \)
- Surrounding temperature, \( T_s = 20°C \)
- Time interval, \( t = 5 \) minutes

2. **Apply Newton's Law of Cooling:**
Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the surrounding temperature. Mathematically, it can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_s) \]
where \( k \) is a positive constant.

3. **Calculate the Cooling Constant \( k \):**
We can rearrange the formula to find \( k \):
\[ \frac{T_1 - T_s}{T_2 - T_s} = e^{-kt} \]
Plugging in the values:
\[ \frac{50 - 20}{40 - 20} = e^{-5k} \]
Simplifying:
\[ \frac{30}{20} = e^{-5k} \implies \frac{3}{2} = e^{-5k} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{3}{2}\right) = -5k \implies k = -\frac{1}{5} \ln\left(\frac{3}{2}\right) \]

4. **Find the Temperature After the Next 5 Minutes:**
Now we need to find the temperature of the body after the next 5 minutes, starting from \( T_2 = 40°C \).
We can use the same formula:
\[ \frac{T_2 - T_s}{T_3 - T_s} = e^{-kt} \]
where \( T_3 \) is the temperature after the next 5 minutes. Plugging in the values:
\[ \frac{40 - 20}{T_3 - 20} = e^{-5k} \]
We already found \( e^{-5k} = \frac{3}{2} \):
\[ \frac{20}{T_3 - 20} = \frac{3}{2} \]
Cross-multiplying gives:
\[ 20 \cdot 2 = 3(T_3 - 20) \]
Simplifying:
\[ 40 = 3T_3 - 60 \]
\[ 3T_3 = 100 \implies T_3 = \frac{100}{3} \approx 33.33°C \]

5. **Conclusion:**
The temperature of the body after the next 5 minutes will be approximately \( 33.33°C \).