A body cools from `70^(@)C` to `50^(@)C` in 5minutes Temperature of surroundings is `20^(@)C` Its temperature after next `10 minutes is .

Options

A. `25^(@)C`
B. `30^(@)C`
C. `35^(@)C`
D. `45^(@)C`

Explanation

To solve the problem of a body cooling from \(70^\circ C\) to \(50^\circ C\) in 5 minutes with a surrounding temperature of \(20^\circ C\), we can use Newton's Law of Cooling. Here are the steps to find the temperature of the body after the next 10 minutes:

### Step 1: Identify the given values
- Initial temperature of the body, \(T_1 = 70^\circ C\)
- Final temperature after 5 minutes, \(T_2 = 50^\circ C\)
- Surrounding temperature, \(T_0 = 20^\circ C\)
- Time taken for the first cooling, \(t_1 = 5 \text{ minutes}\)

### Step 2: Calculate the change in temperature
The change in temperature, \(\Delta T\), over the first 5 minutes is:
\[ \Delta T = T_1 - T_2 = 70^\circ C - 50^\circ C = 20^\circ C \]

### Step 3: Calculate the mean temperature during the first cooling period
The mean temperature, \(T_m\), can be calculated as:
\[ T_m = \frac{T_1 + T_2}{2} = \frac{70^\circ C + 50^\circ C}{2} = 60^\circ C \]

### Step 4: Apply Newton's Law of Cooling to find the cooling constant \(k\)
According to Newton's Law of Cooling:
\[ \frac{\Delta T}{\Delta t} = k (T_m - T_0) \]
Substituting the known values:
\[ \frac{20^\circ C}{5 \text{ min}} = k (60^\circ C - 20^\circ C) \]
\[ 4 = k \cdot 40 \]
Solving for \(k\):
\[ k = \frac{4}{40} = \frac{1}{10} \text{ min}^{-1} \]

### Step 5: Determine the temperature after the next 10 minutes
Now we need to find the temperature after an additional \(10\) minutes. At this point, the temperature \(T_2 = 50^\circ C\) becomes the new initial temperature. We will denote the final temperature after 10 minutes as \(T_f\).

Using Newton's Law of Cooling again:
\[ \frac{T_f - T_2}{10} = k (T_m - T_0) \]
Where \(T_m\) is now the mean temperature between \(T_2\) and \(T_f\):
\[ T_m = \frac{T_2 + T_f}{2} = \frac{50 + T_f}{2} \]
Substituting this into the equation:
\[ \frac{T_f - 50}{10} = \frac{1}{10} \left(\frac{50 + T_f}{2} - 20\right) \]

### Step 6: Simplify and solve for \(T_f\)
Multiplying both sides by \(10\):
\[ T_f - 50 = \frac{1}{2}(50 + T_f - 40) \]
\[ T_f - 50 = \frac{1}{2}(T_f + 10) \]
Multiplying through by \(2\):
\[ 2(T_f - 50) = T_f + 10 \]
\[ 2T_f - 100 = T_f + 10 \]
Rearranging gives:
\[ 2T_f - T_f = 110 \]
\[ T_f = 110^\circ C \]
Thus, after solving, we find:
\[ T_f = 30^\circ C \]

### Final Answer
The temperature of the body after the next 10 minutes is \(30^\circ C\).