A body cools from `60^(@)C` to `50^(@)C` in 10 minutes . If the room temperature is `25^(@)C` and assuming Newton's law of cooling to hold good, the temperature of the body at the end of the next 10 minutes will be

Options

A. `38.5 ^(@)C`
B. `40^(@)C`
C. `42.85^(@)C`
D. `45^(@)C`

Answer

To solve the problem using Newton's law of cooling, we can follow these steps:

### Step 1: Understand the problem
We have a body cooling from an initial temperature of \( T_1 = 60^\circ C \) to a final temperature of \( T_2 = 50^\circ C \) in a time \( t = 10 \) minutes. The room temperature is \( T_r = 25^\circ C \). We need to find the temperature of the body after the next 10 minutes.

### Step 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 temperature and the ambient temperature. Mathematically, it can be expressed as:

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

Where \( k \) is a positive constant.

### Step 3: Calculate the cooling constant \( k \)
In the first 10 minutes, the temperature drops from \( 60^\circ C \) to \( 50^\circ C \). The mean temperature difference can be calculated as:

\[ \text{Mean Temperature Difference} = \frac{T_1 + T_2}{2} - T_r = \frac{60 + 50}{2} - 25 = 30^\circ C \]

The temperature difference during this time is \( 10^\circ C \) (from \( 60^\circ C \) to \( 50^\circ C \)). According to the law, we can write:

\[ \frac{10}{10} = k \cdot 30 \]

This simplifies to:

\[ 1 = k \cdot 30 \implies k = \frac{1}{30} \]

### Step 4: Set up the equation for the next 10 minutes
Now, we need to find the temperature \( T_3 \) after another 10 minutes, starting from \( T_2 = 50^\circ C \). The mean temperature difference in this case will be:

\[ \text{Mean Temperature Difference} = \frac{T_2 + T_3}{2} - T_r = \frac{50 + T_3}{2} - 25 \]

The temperature difference is \( 50 - T_3 \). The equation becomes:

\[ \frac{50 - T_3}{10} = k \left( \frac{50 + T_3}{2} - 25 \right) \]

Substituting \( k = \frac{1}{30} \):

\[ \frac{50 - T_3}{10} = \frac{1}{30} \left( \frac{50 + T_3}{2} - 25 \right) \]

### Step 5: Solve for \( T_3 \)
Multiply both sides by 30 to eliminate the fraction:

\[ 3(50 - T_3) = \frac{50 + T_3}{2} - 25 \]

Expanding and simplifying:

\[ 150 - 3T_3 = \frac{50 + T_3}{2} - 25 \]

Multiply through by 2 to eliminate the fraction:

\[ 300 - 6T_3 = 50 + T_3 - 50 \]

This simplifies to:

\[ 300 - 6T_3 = T_3 \]

Combine like terms:

\[ 300 = 7T_3 \]

Now, divide by 7:

\[ T_3 = \frac{300}{7} \approx 42.86^\circ C \]

### Final Answer
The temperature of the body at the end of the next 10 minutes will be approximately \( 42.86^\circ C \).

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