A body cools from `50^@C " to " 40^@C` in 5 min. If the temperature of the surrounding is `20^@C` , the temperature of the body after the next 5 min would be

Options

A. `36^@C`
B. `35^@C`
C. `33.33^@C`
D. `30^@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.

### Step-by-Step Solution:

**Step 1: Identify the initial conditions.**
- Initial temperature of the body, \( T_1 = 50^\circ C \)
- Final temperature of the body after 5 minutes, \( T_2 = 40^\circ C \)
- Ambient temperature, \( T_a = 20^\circ C \)
- Time interval, \( t = 5 \) minutes

**Step 2: Apply Newton's Law of Cooling for the first interval.**
According to Newton's Law of Cooling:
\[ \frac{dT}{dt} = -k(T - T_a) \]
For the first interval, we can rearrange and integrate:
\[ \frac{T_2 - T_a}{T_1 - T_a} = e^{-kt} \]
Substituting the values:
\[ \frac{40 - 20}{50 - 20} = e^{-5k} \]
This simplifies to:
\[ \frac{20}{30} = e^{-5k} \quad \Rightarrow \quad \frac{2}{3} = e^{-5k} \]

**Step 3: Solve for \( k \).**
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{2}{3}\right) = -5k \]
Thus,
\[ k = -\frac{1}{5} \ln\left(\frac{2}{3}\right) \]

**Step 4: Calculate the temperature after the next 5 minutes.**
Now we need to find the temperature \( T \) after another 5 minutes (total of 10 minutes from the start). The initial temperature for this interval is \( T_2 = 40^\circ C \):
\[ \frac{T - T_a}{T_2 - T_a} = e^{-kt} \]
Substituting the values:
\[ \frac{T - 20}{40 - 20} = e^{-5k} \]
We already found \( e^{-5k} = \frac{2}{3} \):
\[ \frac{T - 20}{20} = \frac{2}{3} \]

**Step 5: Solve for \( T \).**
Cross-multiplying gives:
\[ T - 20 = \frac{2}{3} \times 20 \]
\[ T - 20 = \frac{40}{3} \]
Adding 20 to both sides:
\[ T = 20 + \frac{40}{3} = \frac{60}{3} + \frac{40}{3} = \frac{100}{3} \approx 33.33^\circ C \]

### Final Answer:
The temperature of the body after the next 5 minutes would be approximately \( 33.33^\circ C \).