A body cools from `50^(@)C` to `46^(@)C` in 5 minutes and to `40^(@)C` in the next 10 minutes. The surrounding temperature is :

Options

A. `36^(@)C`
B. `32^(@)C`
C. `30^(@)C`
D. `28^(@)C`

Explanation

To find the surrounding temperature \( T_0 \) using Newton's law of cooling, we can follow these steps:

### Step 1: Define the problem
We have a body cooling from \( 50^\circ C \) to \( 46^\circ C \) in 5 minutes, and then from \( 46^\circ C \) to \( 40^\circ C \) in the next 10 minutes. We need to find the surrounding temperature \( T_0 \).

### Step 2: Apply Newton's Law of Cooling
According to Newton's law of cooling, the rate of change of temperature is proportional to the difference between the temperature of the body and the surrounding temperature:

\[ \frac{dT}{dt} = -k(T - T_0) \]

### Step 3: Set up the first case
For the first case (from \( 50^\circ C \) to \( 46^\circ C \)):
- Initial temperature \( T_1 = 50^\circ C \)
- Final temperature \( T_2 = 46^\circ C \)
- Time \( t_1 = 5 \) minutes

The average temperature \( T_{avg1} \) during this interval is:

\[ T_{avg1} = \frac{T_1 + T_2}{2} = \frac{50 + 46}{2} = 48^\circ C \]

The change in temperature \( \Delta T_1 \) is:

\[ \Delta T_1 = T_1 - T_2 = 50 - 46 = 4^\circ C \]

Using Newton's law of cooling, we can write:

\[ \frac{4}{5} = -k(48 - T_0) \quad \text{(Equation 1)} \]

### Step 4: Set up the second case
For the second case (from \( 46^\circ C \) to \( 40^\circ C \)):
- Initial temperature \( T_3 = 46^\circ C \)
- Final temperature \( T_4 = 40^\circ C \)
- Time \( t_2 = 10 \) minutes

The average temperature \( T_{avg2} \) during this interval is:

\[ T_{avg2} = \frac{T_3 + T_4}{2} = \frac{46 + 40}{2} = 43^\circ C \]

The change in temperature \( \Delta T_2 \) is:

\[ \Delta T_2 = T_3 - T_4 = 46 - 40 = 6^\circ C \]

Using Newton's law of cooling, we can write:

\[ \frac{6}{10} = -k(43 - T_0) \quad \text{(Equation 2)} \]

### Step 5: Simplify the equations
From Equation 1:

\[ \frac{4}{5} = -k(48 - T_0) \implies 4 = -5k(48 - T_0) \implies k(48 - T_0) = -\frac{4}{5} \]

From Equation 2:

\[ \frac{6}{10} = -k(43 - T_0) \implies 6 = -10k(43 - T_0) \implies k(43 - T_0) = -\frac{6}{10} = -\frac{3}{5} \]

### Step 6: Divide the equations
Dividing the two equations:

\[ \frac{\frac{4}{5}}{\frac{6}{10}} = \frac{(48 - T_0)}{(43 - T_0)} \]

This simplifies to:

\[ \frac{4}{5} \cdot \frac{10}{6} = \frac{(48 - T_0)}{(43 - T_0)} \implies \frac{4}{3} = \frac{(48 - T_0)}{(43 - T_0)} \]

### Step 7: Cross-multiply and solve for \( T_0 \)
Cross-multiplying gives:

\[ 4(43 - T_0) = 3(48 - T_0) \]

Expanding both sides:

\[ 172 - 4T_0 = 144 - 3T_0 \]

Rearranging gives:

\[ 172 - 144 = 4T_0 - 3T_0 \implies 28 = T_0 \]

### Final Answer
The surrounding temperature \( T_0 \) is \( 28^\circ C \).

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