The temperature of a body falls from `50^(@)C` to `40^(@)C` in 10 minutes. If the temperature of the surroundings is `20^(@)C` Then temperature of the body after another 10 minutes will be

Options

A. `36.6^(@)C`
B. `33.3^(@)C`
C. `35^(@)C`
D. `30^(@)C`

Answer

To solve the problem of finding the temperature of the body after another 10 minutes, we will use Newton's Law of Cooling. Here’s the step-by-step solution:

### Step 1: Understand the problem and gather the data
- Initial temperature of the body (T1) = 50°C
- Temperature after 10 minutes (T2) = 40°C
- Surrounding temperature (T_s) = 20°C
- Time interval (t) = 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 surrounding temperature. The formula is given by:

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

Where:
- \( T \) is the temperature of the body,
- \( T_s \) is the surrounding temperature,
- \( k \) is a constant.

### Step 3: Calculate the constant \( k \)
Using the data from the first 10 minutes:
- Change in temperature = T1 - T2 = 50°C - 40°C = 10°C
- Time taken = 10 minutes

Using the formula, we can express this as:

\[ \frac{T2 - T1}{t} = -k(T1 - T_s) \]

Substituting the values:

\[ \frac{40 - 50}{10} = -k(50 - 20) \]

This simplifies to:

\[ -1 = -k(30) \]

Thus,

\[ k = \frac{1}{30} \text{ per minute} \]

### Step 4: Find the temperature after another 10 minutes
Now, we need to find the temperature after another 10 minutes, starting from T2 = 40°C. Let the temperature after another 10 minutes be \( T_f \).

Using Newton's Law again:

\[ \frac{T_f - T2}{10} = -k(T2 - T_s) \]

Substituting the values:

\[ \frac{T_f - 40}{10} = -\frac{1}{30}(40 - 20) \]

This simplifies to:

\[ \frac{T_f - 40}{10} = -\frac{1}{30} \times 20 \]

Calculating the right side:

\[ \frac{T_f - 40}{10} = -\frac{20}{30} = -\frac{2}{3} \]

### Step 5: Solve for \( T_f \)
Now, multiply both sides by 10:

\[ T_f - 40 = -\frac{20}{3} \]

Adding 40 to both sides:

\[ T_f = 40 - \frac{20}{3} \]

Converting 40 to a fraction:

\[ T_f = \frac{120}{3} - \frac{20}{3} = \frac{100}{3} \]

Calculating \( \frac{100}{3} \):

\[ T_f \approx 33.33°C \]

### Final Answer
The temperature of the body after another 10 minutes will be approximately **33.33°C**.

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