If a metallic sphere gets cooled from `62^(@)C` to `50^(@)C` in utesmin10 and in the next utes min10 gets cooled to `42^(@)C` , then the temperature of the surroundings is

Options

A. `30^(@)C`
B. `36^(@)C`
C. `26^(@)C`
D. `20^(@)C`

Answer

To solve the problem, we will use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature (temperature of the surroundings).

### Step-by-Step Solution:

1. **Identify the Given Temperatures and Time Intervals:**
- Initial temperature (T1) = 62°C
- Final temperature after 10 minutes (T2) = 50°C
- Next temperature after another 10 minutes (T3) = 42°C
- Time interval (Δt) = 10 minutes for both cooling periods.

2. **Apply Newton's Law of Cooling:**
According to Newton's Law of Cooling:
\[ \frac{dT}{dt} \propto (T - T_0) \]
where \(T_0\) is the surrounding temperature.

3. **Set Up the Equations:**
For the first cooling from 62°C to 50°C:
\[ \frac{T1 - T2}{\Delta t_1} = K \cdot \left(\frac{T1 + T2}{2} - T_0\right) \]
Substituting the values:
\[ \frac{62 - 50}{10} = K \cdot \left(\frac{62 + 50}{2} - T_0\right) \]
Simplifying:
\[ \frac{12}{10} = K \cdot \left(56 - T_0\right) \quad \text{(Equation 1)} \]

For the second cooling from 50°C to 42°C:
\[ \frac{T2 - T3}{\Delta t_2} = K \cdot \left(\frac{T2 + T3}{2} - T_0\right) \]
Substituting the values:
\[ \frac{50 - 42}{10} = K \cdot \left(\frac{50 + 42}{2} - T_0\right) \]
Simplifying:
\[ \frac{8}{10} = K \cdot \left(46 - T_0\right) \quad \text{(Equation 2)} \]

4. **Express K from Both Equations:**
From Equation 1:
\[ K = \frac{12/10}{56 - T_0} \]
From Equation 2:
\[ K = \frac{8/10}{46 - T_0} \]

5. **Set the Two Expressions for K Equal:**
\[ \frac{12/10}{56 - T_0} = \frac{8/10}{46 - T_0} \]

6. **Cross Multiply to Solve for T0:**
\[ 12(46 - T_0) = 8(56 - T_0) \]
Expanding both sides:
\[ 552 - 12T_0 = 448 - 8T_0 \]
Rearranging gives:
\[ 552 - 448 = 12T_0 - 8T_0 \]
\[ 104 = 4T_0 \]
\[ T_0 = 26°C \]

### Conclusion:
The temperature of the surroundings is **26°C**.