A body cools in 10 minutes from `60^(@)C` to `40^(@)C`. What is the temperature of the body after next 20 minutes? The temperatuer fo surroundings is `10^(@)C`

Options

A. `14^(@)C`
B. `30^(@)C`
C. `20.8^(@)C`
D. `14.36^(@)C`

Answer

To solve the problem of determining the temperature of a body after it cools for an additional 20 minutes, we can use Newton's Law of Cooling. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We know that the body cools from 60°C to 40°C in 10 minutes, and we want to find the temperature after an additional 20 minutes, given that the surrounding temperature is 10°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 surrounding temperature. The formula can be expressed as:

\[ \frac{\Delta T}{\Delta t} = k (T - T_0) \]

Where:
- \( \Delta T \) is the change in temperature,
- \( \Delta t \) is the time interval,
- \( k \) is a constant,
- \( T \) is the temperature of the body,
- \( T_0 \) is the surrounding temperature.

### Step 3: Set Up the First Cooling Period
For the first cooling period (from 60°C to 40°C in 10 minutes):

- Initial temperature \( T_1 = 60°C \)
- Final temperature \( T_2 = 40°C \)
- Time \( \Delta t_1 = 10 \) minutes
- Surrounding temperature \( T_0 = 10°C \)

Using the formula:

\[ \frac{60 - 40}{10} = k \left(\frac{60 + 40}{2} - 10\right) \]

Calculating the left side:

\[ \frac{20}{10} = 2 \]

Calculating the right side:

\[ k \left(\frac{100}{2} - 10\right) = k(50 - 10) = k \cdot 40 \]

Setting the two sides equal:

\[ 2 = k \cdot 40 \implies k = \frac{2}{40} = \frac{1}{20} \]

### Step 4: Set Up the Second Cooling Period
Now, we consider the next cooling period (from 40°C to some final temperature \( T \) in 20 minutes):

- Initial temperature \( T_1 = 40°C \)
- Final temperature \( T_2 = T \)
- Time \( \Delta t_2 = 20 \) minutes

Using Newton's law again:

\[ \frac{40 - T}{20} = k \left(\frac{40 + T}{2} - 10\right) \]

Substituting \( k = \frac{1}{20} \):

\[ \frac{40 - T}{20} = \frac{1}{20} \left(\frac{40 + T}{2} - 10\right) \]

### Step 5: Simplify and Solve for \( T \)
Multiplying both sides by 20 to eliminate the fraction:

\[ 40 - T = \frac{40 + T}{2} - 10 \]

Multiply through by 2 to eliminate the fraction:

\[ 2(40 - T) = 40 + T - 20 \]

This simplifies to:

\[ 80 - 2T = 20 + T \]

Rearranging gives:

\[ 80 - 20 = 2T + T \implies 60 = 3T \implies T = 20°C \]

### Conclusion
The temperature of the body after the next 20 minutes is **20°C**.

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