A liquid cools from 70 °C to 60 °C in 5 minutes. If the temperature of the surrounding is constant at 30 °C, then the time taken by the liquid to cool from 60 °C to 50 °C is

Options

A. 5 minutes
B. 10 minutes
C. 7 minutes
D. 8 minutes

Answer

To solve the problem of how long it takes for a liquid to cool from 60 °C to 50 °C, we will apply Newton's Law of Cooling. Here are the steps to arrive at the solution:

### Step 1: Understand the Problem
We know that the liquid cools from 70 °C to 60 °C in 5 minutes, and the surrounding temperature is constant at 30 °C. We need to find the time taken to cool from 60 °C to 50 °C.

### Step 2: Apply Newton's Law of Cooling
According to Newton's Law of Cooling, the rate of cooling is proportional to the difference in temperature between the body and its surroundings. Mathematically, this can be expressed as:

\[ \frac{dT}{dt} \propto (T_b - T_s) \]

Where:
- \( T_b \) = temperature of the body
- \( T_s \) = temperature of the surroundings

### Step 3: Set Up the Equation for the First Cooling Interval
For the first cooling interval (from 70 °C to 60 °C):

- Initial temperature \( T_{b1} = 70 \) °C
- Final temperature \( T_{b2} = 60 \) °C
- Surrounding temperature \( T_s = 30 \) °C
- Time taken \( \Delta t_1 = 5 \) minutes

Using the average temperature during this interval:

\[ T_{avg1} = \frac{T_{b1} + T_{b2}}{2} = \frac{70 + 60}{2} = 65 \text{ °C} \]

The difference in temperature is:

\[ \Delta T_1 = T_{b1} - T_s = 70 - 30 = 40 \text{ °C} \]

### Step 4: Calculate the Cooling Constant \( k \)
From the first cooling interval, we can express the relationship as:

\[ \frac{\Delta T_1}{\Delta t_1} = k \cdot (T_{avg1} - T_s) \]

Substituting the values:

\[ \frac{40}{5} = k \cdot (65 - 30) \]

This simplifies to:

\[ 8 = k \cdot 35 \]

Thus, we find:

\[ k = \frac{8}{35} \]

### Step 5: Set Up the Equation for the Second Cooling Interval
Now we analyze the second cooling interval (from 60 °C to 50 °C):

- Initial temperature \( T_{b1} = 60 \) °C
- Final temperature \( T_{b2} = 50 \) °C
- Time taken \( \Delta t_2 \) (unknown)

Using the average temperature during this interval:

\[ T_{avg2} = \frac{T_{b1} + T_{b2}}{2} = \frac{60 + 50}{2} = 55 \text{ °C} \]

The difference in temperature is:

\[ \Delta T_2 = T_{b1} - T_s = 60 - 30 = 30 \text{ °C} \]

### Step 6: Calculate the Time for the Second Cooling Interval
Using the same relationship:

\[ \frac{\Delta T_2}{\Delta t_2} = k \cdot (T_{avg2} - T_s) \]

Substituting the values:

\[ \frac{30}{\Delta t_2} = \frac{8}{35} \cdot (55 - 30) \]

This simplifies to:

\[ \frac{30}{\Delta t_2} = \frac{8}{35} \cdot 25 \]

Calculating the right side:

\[ \frac{30}{\Delta t_2} = \frac{200}{35} \]

Cross-multiplying gives:

\[ 30 \cdot 35 = 200 \cdot \Delta t_2 \]

Thus:

\[ \Delta t_2 = \frac{30 \cdot 35}{200} = \frac{1050}{200} = 5.25 \text{ minutes} \]

### Conclusion
The time taken by the liquid to cool from 60 °C to 50 °C is approximately **5.25 minutes**.