A body cools in 3 minute from 90°C to 80°C. The temperature reduce to 70°C in next (If temperature of surroundings is 20°C)

Options

A. 2.54 minutes
B. 2 minutes
C. 6 minutes
D. 3.54 minutes

Answer

To solve the problem of how long it takes for a body to cool from 80°C to 70°C, we will use Newton's Law of Cooling. Here are the steps to find the solution:

### Step 1: Understand the cooling process
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.

### Step 2: Set up the equation for the first cooling phase
We know that the body cools from 90°C to 80°C in 3 minutes. We can express this using Newton's Law of Cooling:

\[ \frac{T_f - T_i}{t} = K \left( \frac{T_f + T_i}{2} - T_0 \right) \]

Where:
- \( T_f = 80°C \) (final temperature)
- \( T_i = 90°C \) (initial temperature)
- \( t = 3 \) minutes (time taken)
- \( T_0 = 20°C \) (surrounding temperature)
- \( K \) is the cooling constant.

### Step 3: Substitute values into the equation
Substituting the values into the equation:

\[ \frac{80 - 90}{3} = K \left( \frac{80 + 90}{2} - 20 \right) \]

Calculating the left side:

\[ \frac{-10}{3} = K \left( \frac{170}{2} - 20 \right) \]
\[ \frac{-10}{3} = K \left( 85 - 20 \right) \]
\[ \frac{-10}{3} = K \times 65 \]

### Step 4: Solve for K
Now, we can solve for \( K \):

\[ K = \frac{-10/3}{65} = \frac{-10}{195} = -\frac{2}{39} \]

### Step 5: Set up the equation for the second cooling phase
Next, we need to find the time it takes for the temperature to drop from 80°C to 70°C. We will use the same formula:

\[ \frac{T_f - T_i}{t} = K \left( \frac{T_f + T_i}{2} - T_0 \right) \]

Where:
- \( T_f = 70°C \)
- \( T_i = 80°C \)

Substituting these values into the equation:

\[ \frac{70 - 80}{t} = K \left( \frac{70 + 80}{2} - 20 \right) \]

Calculating the left side:

\[ \frac{-10}{t} = K \left( \frac{150}{2} - 20 \right) \]
\[ \frac{-10}{t} = K \left( 75 - 20 \right) \]
\[ \frac{-10}{t} = K \times 55 \]

### Step 6: Substitute the value of K
Substituting the value of \( K = -\frac{2}{39} \):

\[ \frac{-10}{t} = -\frac{2}{39} \times 55 \]

### Step 7: Solve for t
Now, we can solve for \( t \):

\[ \frac{-10}{t} = -\frac{110}{39} \]

Cross-multiplying gives:

\[ -10 \times 39 = -110 \times t \]
\[ t = \frac{10 \times 39}{110} \]
\[ t = \frac{390}{110} = 3.545 \text{ minutes} \approx 3.54 \text{ minutes} \]

### Final Answer
The time taken for the temperature to reduce from 80°C to 70°C is approximately **3.54 minutes**.

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