A black body calorimeter filled with hot water cools from `60^(@)C` to `50^(@)C` in `4 min` and `40^(@)C` to `30^(@)C` in `8min`. The approximate temperature of surrounding is :

Options

A. `10^(@)`
B. `15^(@)C`
C. `20^(@)C`
D. `25^(@)C`

Answer

To find the approximate temperature of the surrounding using the given data, we can apply Newton's Law of Cooling. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We are given two cooling intervals:
1. From \(60^\circ C\) to \(50^\circ C\) in \(4\) minutes.
2. From \(40^\circ C\) to \(30^\circ C\) in \(8\) minutes.

We need to find the surrounding temperature (\(\theta_n\)).

### Step 2: Apply Newton's Law of Cooling
According to Newton's Law of Cooling:
\[ \frac{\Delta T}{t} = k \left( \frac{T_1 + T_2}{2} - \theta_n \right) \]
where:
- \(\Delta T\) is the change in temperature,
- \(t\) is the time taken,
- \(k\) is a constant,
- \(T_1\) and \(T_2\) are the initial and final temperatures.

### Step 3: Set Up the Equations
For the first cooling interval (from \(60^\circ C\) to \(50^\circ C\)):
\[ \frac{60 - 50}{4} = k \left( \frac{60 + 50}{2} - \theta_n \right) \]
This simplifies to:
\[ \frac{10}{4} = k \left( 55 - \theta_n \right) \]
Thus, we have:
\[ 2.5 = k(55 - \theta_n) \quad \text{(Equation 1)} \]

For the second cooling interval (from \(40^\circ C\) to \(30^\circ C\)):
\[ \frac{40 - 30}{8} = k \left( \frac{40 + 30}{2} - \theta_n \right) \]
This simplifies to:
\[ \frac{10}{8} = k \left( 35 - \theta_n \right) \]
Thus, we have:
\[ 1.25 = k(35 - \theta_n) \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now, we will divide Equation 1 by Equation 2:
\[ \frac{2.5}{1.25} = \frac{k(55 - \theta_n)}{k(35 - \theta_n)} \]
This simplifies to:
\[ 2 = \frac{55 - \theta_n}{35 - \theta_n} \]

Cross-multiplying gives:
\[ 2(35 - \theta_n) = 55 - \theta_n \]
Expanding this results in:
\[ 70 - 2\theta_n = 55 - \theta_n \]

### Step 5: Isolate \(\theta_n\)
Rearranging the equation:
\[ 70 - 55 = 2\theta_n - \theta_n \]
This simplifies to:
\[ 15 = \theta_n \]

### Conclusion
The approximate temperature of the surrounding is:
\[ \theta_n = 15^\circ C \]