Hot water cools from `60^@C` to `50^@C` in the first 10 min and to `42^@C` in the next 10 min. The temperature of the surrounding is

Options

A. `5^@C`
B. `10^@C`
C. `15^@C`
D. `20^@C`

Explanation

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 temperature and the ambient temperature (temperature of the surroundings).

### Step-by-Step Solution:

1. **Identify the temperatures and time intervals:**
- Initial temperature \( T_1 = 60^\circ C \)
- Temperature after 10 minutes \( T_2 = 50^\circ C \)
- Temperature after another 10 minutes \( T_3 = 42^\circ C \)
- Time intervals: \( t_1 = 10 \) minutes for the first cooling, \( t_2 = 10 \) minutes for the second cooling.

2. **Apply Newton's Law of Cooling:**
For the first interval (from \( 60^\circ C \) to \( 50^\circ C \)):
\[ \frac{T_1 - T_2}{t_1} \propto \left( \frac{T_1 + T_2}{2} - T_s \right) \]
This gives us:
\[ \frac{60 - 50}{10} \propto \left( \frac{60 + 50}{2} - T_s \right) \]
Simplifying:
\[ 1 \propto \left( 55 - T_s \right) \quad \text{(Equation 1)} \]

3. **For the second interval (from \( 50^\circ C \) to \( 42^\circ C \)):**
\[ \frac{T_2 - T_3}{t_2} \propto \left( \frac{T_2 + T_3}{2} - T_s \right) \]
This gives us:
\[ \frac{50 - 42}{10} \propto \left( \frac{50 + 42}{2} - T_s \right) \]
Simplifying:
\[ \frac{8}{10} \propto \left( 46 - T_s \right) \quad \text{(Equation 2)} \]

4. **Set up the proportionality equations:**
From Equation 1:
\[ 1 = k(55 - T_s) \quad \text{(where \( k \) is a constant)} \]
From Equation 2:
\[ \frac{8}{10} = k(46 - T_s) \]

5. **Eliminate \( k \) by dividing the equations:**
\[ \frac{1}{\frac{8}{10}} = \frac{55 - T_s}{46 - T_s} \]
Simplifying gives:
\[ \frac{10}{8} = \frac{55 - T_s}{46 - T_s} \]
Cross-multiplying:
\[ 10(46 - T_s) = 8(55 - T_s) \]
Expanding both sides:
\[ 460 - 10T_s = 440 - 8T_s \]

6. **Rearranging the equation:**
\[ 460 - 440 = 10T_s - 8T_s \]
\[ 20 = 2T_s \]
\[ T_s = 10^\circ C \]

### Final Answer:
The temperature of the surrounding is \( T_s = 10^\circ C \).