A body cools from `50^@C` to `49.9^@C` in 5 s. How long will it take to cool from `40^@C` to `39.9^@C`? Assume the temperature of surroundings to be `30^@C` and Newton's law of cooling to be valid:

Options

A. 2.5 s
B. 10 s
C. 20 s
D. 5 s

Answer

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

### 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 temperature and the ambient temperature. Mathematically, it can be expressed as:
\[ \frac{d\theta}{dt} = -k(\theta - \theta_0) \]
where:
- \(\theta\) is the temperature of the body,
- \(\theta_0\) is the ambient temperature,
- \(k\) is a positive constant (thermal conductivity).

### Step 2: Set Up the First Cooling Scenario
In the first scenario, the body cools from \(50^\circ C\) to \(49.9^\circ C\) in \(5\) seconds. We can denote:
- \(\theta_1 = 50^\circ C\)
- \(\theta_2 = 49.9^\circ C\)
- \(\theta_0 = 30^\circ C\)
- \(T_1 = 5\) seconds

Using Newton's law of cooling:
\[ \frac{\theta_1 - \theta_2}{T_1} = k \left(\frac{\theta_1 + \theta_2}{2} - \theta_0\right) \]
Substituting the values:
\[ \frac{50 - 49.9}{5} = k \left(\frac{50 + 49.9}{2} - 30\right) \]
This simplifies to:
\[ \frac{0.1}{5} = k \left(19.95 - 30\right) \]
\[ 0.02 = k \cdot 19.95 \]
Thus, we can express \(k\):
\[ k = \frac{0.02}{19.95} \]

### Step 3: Set Up the Second Cooling Scenario
In the second scenario, the body cools from \(40^\circ C\) to \(39.9^\circ C\). We denote:
- \(\theta_3 = 40^\circ C\)
- \(\theta_4 = 39.9^\circ C\)
- \(T_2\) is what we want to find.

Using Newton's law of cooling again:
\[ \frac{\theta_3 - \theta_4}{T_2} = k \left(\frac{\theta_3 + \theta_4}{2} - \theta_0\right) \]
Substituting the values:
\[ \frac{40 - 39.9}{T_2} = k \left(\frac{40 + 39.9}{2} - 30\right) \]
This simplifies to:
\[ \frac{0.1}{T_2} = k \cdot 9.95 \]

### Step 4: Relate the Two Scenarios
Now we can relate the two scenarios:
From the first scenario:
\[ k = \frac{0.02}{19.95} \]
From the second scenario:
\[ \frac{0.1}{5} = k \cdot 19.95 \]
And:
\[ \frac{0.1}{T_2} = k \cdot 9.95 \]
Dividing the two equations:
\[ \frac{0.1 / 5}{0.1 / T_2} = \frac{k \cdot 19.95}{k \cdot 9.95} \]
This simplifies to:
\[ \frac{T_2}{5} = \frac{19.95}{9.95} \]
Thus:
\[ T_2 = 5 \cdot \frac{19.95}{9.95} \]
Calculating \(T_2\):
\[ T_2 \approx 10 \text{ seconds} \]

### Final Answer
It will take approximately \(10\) seconds for the body to cool from \(40^\circ C\) to \(39.9^\circ C\).

---