A cup of tea cools from `80^(circ) C` to `60^(circ) C` in one min. The ambient temperature is `30^(circ) C`. In cooling from `60^(circ) C` to ` 0^(circ) C` how much time (in s) will it take?

Answer

To solve the problem of how long it takes for a cup of tea to cool from 60°C to 0°C, we can use Newton's Law of Cooling. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the scenario
We know that the cup of tea cools from 80°C to 60°C in 1 minute (60 seconds). The ambient temperature (room temperature) is 30°C. We need to find out how long it will take for the tea to cool from 60°C to 0°C.

### Step 2: Apply 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:

\[ \frac{dT}{dt} = -k(T - T_0) \]

Where:
- \( T \) is the temperature of the object,
- \( T_0 \) is the ambient temperature,
- \( k \) is a constant.

### Step 3: Set up the first cooling scenario
For the first cooling scenario (from 80°C to 60°C):
- Initial temperature \( T_1 = 80°C \)
- Final temperature \( T_2 = 60°C \)
- Time taken \( t_1 = 60 \) seconds
- Ambient temperature \( T_0 = 30°C \)

Using the formula, we can express this as:

\[ \frac{80 - 30}{60} = k \left( \frac{80 + 60}{2} - 30 \right) \]

Calculating the left side:

\[ \frac{50}{60} = \frac{5}{6} \]

Calculating the right side:

\[ \frac{80 + 60}{2} = 70 \quad \Rightarrow \quad 70 - 30 = 40 \]

Thus, we have:

\[ \frac{5}{6} = k \cdot 40 \]

From this, we can solve for \( k \):

\[ k = \frac{5}{6 \cdot 40} = \frac{5}{240} = \frac{1}{48} \]

### Step 4: Set up the second cooling scenario
Now, for the second cooling scenario (from 60°C to 0°C):
- Initial temperature \( T_1' = 60°C \)
- Final temperature \( T_2' = 0°C \)
- Time taken \( t_2 \) (which we need to find)
- Ambient temperature \( T_0 = 30°C \)

Using the same formula:

\[ \frac{60 - 30}{t_2} = k \left( \frac{60 + 0}{2} - 30 \right) \]

Calculating the left side:

\[ \frac{30}{t_2} \]

Calculating the right side:

\[ \frac{60 + 0}{2} = 30 \quad \Rightarrow \quad 30 - 30 = 0 \]

Thus, we have:

\[ \frac{30}{t_2} = k \cdot 0 \]

Since the right side equals zero, this implies that:

\[ \frac{30}{t_2} = 0 \quad \Rightarrow \quad t_2 \rightarrow \infty \]

### Conclusion
The time taken for the tea to cool from 60°C to 0°C is infinite. This is because the temperature difference between the tea and the ambient temperature becomes very small, making the cooling process extremely slow.