A cup of tea cools from `65.5^@`C to `62.55^@C` in one minute is a room at `225 .^@C.` How long will the same cup of tea take to cool from `46.5^@C` to `40.5^@C` in the same room ? (Choose the nearest value in min).

Options

A. 4 min
B. 2 min
C. 1 min
D. 3 min

Answer

To solve the problem of how long it will take for a cup of tea to cool from 46.5°C to 40.5°C in a room at 22.5°C, we can use Newton's Law of Cooling. Here’s a step-by-step solution:

### Step 1: Understand the problem
We know that the temperature of the tea cools from 65.5°C to 62.5°C in 1 minute in a room at 22.5°C. We need to find out how long it will take for the tea to cool from 46.5°C to 40.5°C in the same room.

### Step 2: Apply Newton's Law of Cooling
According to Newton's Law of Cooling, the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature.

The formula can be expressed as:
\[ \frac{dT}{dt} = -k(T - T_{room}) \]
Where:
- \(T\) is the temperature of the object (tea),
- \(T_{room}\) is the ambient temperature (22.5°C),
- \(k\) is a positive constant.

### Step 3: Set up the equations for the first cooling scenario
For the first scenario:
- Initial temperature \(T_1 = 65.5°C\)
- Final temperature \(T_2 = 62.5°C\)
- Room temperature \(T_{room} = 22.5°C\)
- Time taken \(t_1 = 1 \text{ minute}\)

Using the formula:
\[ \frac{T_1 - T_2}{t_1} = -k \left( \frac{T_1 + T_2}{2} - T_{room} \right) \]
Substituting the values:
\[ \frac{65.5 - 62.5}{1} = -k \left( \frac{65.5 + 62.5}{2} - 22.5 \right) \]
Calculating the left side:
\[ 3 = -k \left( \frac{128}{2} - 22.5 \right) \]
\[ 3 = -k (64 - 22.5) = -k (41.5) \]
Thus, we have:
\[ k = -\frac{3}{41.5} \]

### Step 4: Set up the equations for the second cooling scenario
For the second scenario:
- Initial temperature \(T_1 = 46.5°C\)
- Final temperature \(T_2 = 40.5°C\)
- Room temperature \(T_{room} = 22.5°C\)
- Time taken \(t_2\) (unknown)

Using the same formula:
\[ \frac{T_1 - T_2}{t_2} = -k \left( \frac{T_1 + T_2}{2} - T_{room} \right) \]
Substituting the values:
\[ \frac{46.5 - 40.5}{t_2} = -k \left( \frac{46.5 + 40.5}{2} - 22.5 \right) \]
Calculating the left side:
\[ \frac{6}{t_2} = -k \left( \frac{87}{2} - 22.5 \right) \]
Calculating further:
\[ \frac{6}{t_2} = -k (43.5 - 22.5) = -k (21) \]

### Step 5: Relate the two scenarios
Now we have two equations:
1. \(3 = -k (41.5)\)
2. \(\frac{6}{t_2} = -k (21)\)

From the first equation, we can express \(k\):
\[ k = -\frac{3}{41.5} \]

Substituting \(k\) into the second equation:
\[ \frac{6}{t_2} = -\left(-\frac{3}{41.5}\right) (21) \]
\[ \frac{6}{t_2} = \frac{3 \times 21}{41.5} \]
\[ \frac{6}{t_2} = \frac{63}{41.5} \]
Now, solving for \(t_2\):
\[ t_2 = \frac{6 \times 41.5}{63} \]
Calculating \(t_2\):
\[ t_2 \approx \frac{249}{63} \approx 3.95 \text{ minutes} \]

### Step 6: Round to the nearest minute
Thus, the time taken for the tea to cool from 46.5°C to 40.5°C is approximately **4 minutes**.