A cup of tea cools from `80^(@)C" to "60^(@)C` in 40 seconds. The ambient temperature is `30^(@)C`. In cooling from `60^(@)C" to "50^(@)C`, it will take time:

Options

A. 35 s
B. 30 s
C. 32 s
D. 48 s

Answer

To solve the problem of how long it takes for a cup of tea to cool from 60°C to 50°C, we can use Newton's Law of Cooling. Let's break down the solution step by step:

### Step 1: Understand the Problem
We know the following:
- The initial temperature of the tea (T1) = 80°C
- The final temperature after 40 seconds (T2) = 60°C
- The ambient temperature (Ts) = 30°C
- The time taken to cool from 80°C to 60°C (Δt1) = 40 seconds

We need to find the time taken to cool from 60°C to 50°C (Δt2).

### Step 2: Apply Newton's Law of Cooling
Newton's Law of Cooling states:
\[ \frac{dT}{dt} = -k(T - Ts) \]
Where:
- \(T\) is the temperature of the object,
- \(Ts\) is the surrounding temperature,
- \(k\) is a constant.

For a finite change in temperature, we can rearrange this to:
\[ \frac{T_i - T_f}{\Delta t} = k \cdot (T_{avg} - Ts) \]

### Step 3: Set Up the First Equation
For the first cooling process (from 80°C to 60°C):
- \(T_i = 80°C\)
- \(T_f = 60°C\)
- \(T_{avg} = \frac{80 + 60}{2} = 70°C\)

Substituting these values into the equation:
\[ \frac{80 - 60}{40} = k \cdot (70 - 30) \]
\[ \frac{20}{40} = k \cdot 40 \]
\[ 0.5 = 40k \]
\[ k = \frac{0.5}{40} = \frac{1}{80} \]

### Step 4: Set Up the Second Equation
For the second cooling process (from 60°C to 50°C):
- \(T_i = 60°C\)
- \(T_f = 50°C\)
- \(T_{avg} = \frac{60 + 50}{2} = 55°C\)

Substituting these values into the equation:
\[ \frac{60 - 50}{\Delta t_2} = k \cdot (55 - 30) \]
\[ \frac{10}{\Delta t_2} = k \cdot 25 \]

### Step 5: Substitute the Value of k
Now, substituting \(k = \frac{1}{80}\) into the second equation:
\[ \frac{10}{\Delta t_2} = \frac{1}{80} \cdot 25 \]
\[ \frac{10}{\Delta t_2} = \frac{25}{80} \]
\[ \Delta t_2 = \frac{10 \cdot 80}{25} \]
\[ \Delta t_2 = \frac{800}{25} = 32 \text{ seconds} \]

### Final Answer
The time taken for the tea to cool from 60°C to 50°C is **32 seconds**.

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