A metallic sphere cools from 60°C to 52°C in 10 minutes. If the surrounding temperature is 16° C, then the temperature of sphere after the next 10 minutes is close to :

Options

A. 44°C
B. `45^(@)C`
C. `49^(@)C`
D. `51^(@)C`

Answer

To solve the problem of determining the temperature of a metallic sphere after it cools for an additional 10 minutes, we can use Newton's Law of Cooling. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
- Initial temperature of the sphere, \( T_1 = 60^\circ C \)
- Temperature after 10 minutes, \( T_2 = 52^\circ C \)
- Surrounding temperature, \( T_0 = 16^\circ C \)
- Time interval for the first cooling, \( t_1 = 10 \) minutes

### 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 surrounding temperature. The formula can be expressed as:

\[ \frac{T_1 - T_2}{t_1} = k \left( \frac{T_1 + T_2}{2} - T_0 \right) \]

### Step 3: Substitute Known Values
Substituting the known values into the formula:

\[ \frac{60 - 52}{10} = k \left( \frac{60 + 52}{2} - 16 \right) \]

Calculating the left side:

\[ \frac{8}{10} = 0.8 \]

Calculating the right side:

\[ \frac{60 + 52}{2} = \frac{112}{2} = 56 \]

Thus, we have:

\[ 0.8 = k (56 - 16) \]

This simplifies to:

\[ 0.8 = k \cdot 40 \]

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

\[ k = \frac{0.8}{40} = \frac{1}{50} \]

### Step 5: Calculate the Temperature After the Next 10 Minutes
Now we need to find the temperature of the sphere after another 10 minutes. We will denote this temperature as \( T \). Again applying Newton's Law of Cooling for the next interval:

\[ \frac{T_2 - T}{t_2} = k \left( \frac{T_2 + T}{2} - T_0 \right) \]

Where \( t_2 = 10 \) minutes. Substituting the known values:

\[ \frac{52 - T}{10} = \frac{1}{50} \left( \frac{52 + T}{2} - 16 \right) \]

### Step 6: Simplify and Solve for \( T \)
Multiply both sides by 50 to eliminate the fraction:

\[ 5(52 - T) = \frac{52 + T}{2} - 16 \]

Expanding and simplifying:

\[ 260 - 5T = \frac{52 + T}{2} - 16 \]

Multiply through by 2 to eliminate the fraction:

\[ 520 - 10T = 52 + T - 32 \]

Combine like terms:

\[ 520 - 10T = 20 + T \]

Rearranging gives:

\[ 520 - 20 = 11T \]

Thus:

\[ 500 = 11T \]

### Step 7: Calculate \( T \)
Finally, solving for \( T \):

\[ T = \frac{500}{11} \approx 45.45^\circ C \]

### Conclusion
The temperature of the sphere after the next 10 minutes is approximately \( 45.45^\circ C \), which can be rounded to \( 45^\circ C \).