A liquid cools down from `70^(@)C` to `60^(@)C` in 5 minutes. The time taken to cool it from `60^(@)C` to `50^(@)C` will be

Options

A. 5 minutes
B. lessser than 5 minutes
C. greater than 5 minutes
D. Lesser or greater than 5 minutes depending upon the density of the liquid.

Answer

To solve the problem of how long it takes for a liquid to cool from \(60^\circ C\) to \(50^\circ C\), given that it cools from \(70^\circ C\) to \(60^\circ C\) in 5 minutes, we can use Newton's Law of Cooling. Here’s a step-by-step breakdown of the 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 temperature and the surrounding temperature. Mathematically, it can be expressed as:
\[ \frac{d\theta}{dt} = -k(\theta - \theta_s) \]
where:
- \(\theta\) is the temperature of the object,
- \(\theta_s\) is the surrounding temperature,
- \(k\) is a constant.

### Step 2: Set Up the Equations
For the first cooling process (from \(70^\circ C\) to \(60^\circ C\)):
- Initial temperature \(\theta_1 = 70^\circ C\)
- Final temperature \(\theta_2 = 60^\circ C\)
- Time taken \(t_1 = 5 \text{ minutes}\)

Using the average temperature:
\[ \text{Average} = \frac{70 + 60}{2} = 65^\circ C \]
Assuming the surrounding temperature \(\theta_s = 25^\circ C\), we can write:
\[ \frac{70 - 60}{5} = k \left(65 - 25\right) \]
This simplifies to:
\[ \frac{10}{5} = k \cdot 40 \implies 2 = 40k \implies k = \frac{1}{20} \]

### Step 3: Set Up the Second Cooling Process
For the second cooling process (from \(60^\circ C\) to \(50^\circ C\)):
- Initial temperature \(\theta_1 = 60^\circ C\)
- Final temperature \(\theta_2 = 50^\circ C\)
- Let the time taken be \(t_2\).

Using the average temperature:
\[ \text{Average} = \frac{60 + 50}{2} = 55^\circ C \]
We can write:
\[ \frac{60 - 50}{t_2} = k \left(55 - 25\right) \]
This simplifies to:
\[ \frac{10}{t_2} = k \cdot 30 \]

### Step 4: Substitute the Value of \(k\)
Substituting \(k = \frac{1}{20}\):
\[ \frac{10}{t_2} = \frac{1}{20} \cdot 30 \implies \frac{10}{t_2} = \frac{30}{20} \implies \frac{10}{t_2} = 1.5 \]

### Step 5: Solve for \(t_2\)
Cross-multiplying gives:
\[ 10 = 1.5t_2 \implies t_2 = \frac{10}{1.5} = \frac{20}{3} \text{ minutes} \approx 6.67 \text{ minutes} \]

### Conclusion
The time taken to cool the liquid from \(60^\circ C\) to \(50^\circ C\) is approximately \(6.67\) minutes, which is greater than \(5\) minutes.

---