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 of a body cooling from \(50^\circ C\) to \(40^\circ C\) in 5 minutes with a surrounding temperature of \(20^\circ C\), we will use Newton's Law of Cooling. Here’s a step-by-step solution:

### Step 1: Understand 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 ambient temperature. Mathematically, it 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,
- \(k\) is a positive constant.

### Step 2: Set Up the Initial Conditions
In the first 5 minutes, the body cools from \(T_i = 50^\circ C\) to \(T_f = 40^\circ C\). The surrounding temperature \(T_s = 20^\circ C\).

### Step 3: Calculate the Cooling Constant \(k\)
Using the average temperature during the first cooling period, we can express it as:

\[ \text{Average Temperature} = \frac{T_i + T_f}{2} = \frac{50 + 40}{2} = 45^\circ C \]

Using the formula for the change in temperature over time, we can write:

\[ \frac{T_i - T_f}{\Delta t} = k \left( \frac{T_i + T_f}{2} - T_s \right) \]

Substituting the known values:

\[ \frac{50 - 40}{5} = k \left( 45 - 20 \right) \]

This simplifies to:

\[ 2 = k \cdot 25 \]

Solving for \(k\):

\[ k = \frac{2}{25} \]

### Step 4: Calculate the Temperature After the Next 5 Minutes
Now, we need to find the temperature of the body after another 5 minutes, starting from \(T_i = 40^\circ C\) and letting the final temperature be \(T'\).

Using the same formula:

\[ \frac{T_i - T'}{\Delta t} = k \left( \frac{T_i + T'}{2} - T_s \right) \]

Substituting the values:

\[ \frac{40 - T'}{5} = \frac{2}{25} \left( \frac{40 + T'}{2} - 20 \right) \]

### Step 5: Simplifying the Equation
Multiply both sides by 5:

\[ 40 - T' = \frac{2}{25} \cdot 5 \left( \frac{40 + T'}{2} - 20 \right) \]

This simplifies to:

\[ 40 - T' = \frac{2}{5} \left( \frac{40 + T'}{2} - 20 \right) \]

Multiply out the right side:

\[ 40 - T' = \frac{2}{5} \left( 20 + \frac{T'}{2} - 20 \right) \]

This reduces to:

\[ 40 - T' = \frac{2}{5} \cdot \frac{T'}{2} \]

\[ 40 - T' = \frac{T'}{5} \]

### Step 6: Solve for \(T'\)
Rearranging gives:

\[ 40 = T' + \frac{T'}{5} \]

Combining terms:

\[ 40 = \frac{5T' + T'}{5} = \frac{6T'}{5} \]

Multiplying through by 5:

\[ 200 = 6T' \]

Thus:

\[ T' = \frac{200}{6} = 33.33^\circ C \]

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

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