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. `20^@C`
B. `30^@C`
C. `15^@C`
D. `10^@C`

Answer

To solve the problem of determining the temperature of the surroundings using Newton's Law of Cooling, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the temperature changes**:
- The hot water cools from \(60^\circ C\) to \(50^\circ C\) in the first 10 minutes.
- Then it cools from \(50^\circ C\) to \(42^\circ C\) in the next 10 minutes.

2. **Calculate the average temperatures**:
- For the first interval (from \(60^\circ C\) to \(50^\circ C\)):
\[ T_{avg1} = \frac{60 + 50}{2} = \frac{110}{2} = 55^\circ C \]
- For the second interval (from \(50^\circ C\) to \(42^\circ C\)):
\[ T_{avg2} = \frac{50 + 42}{2} = \frac{92}{2} = 46^\circ C \]

3. **Apply Newton's Law of Cooling**:
- According to Newton's Law of Cooling, the rate of change of temperature is proportional to the difference between the temperature of the body and the surrounding temperature.
- Let \(T_s\) be the surrounding temperature.
- The temperature differences for the two intervals are:
- For the first interval:
\[ \Delta T_1 = 60 - 50 = 10^\circ C \]
- For the second interval:
\[ \Delta T_2 = 50 - 42 = 8^\circ C \]

4. **Set up the proportionality**:
- From Newton's Law of Cooling, we can write:
\[ \frac{\Delta T_1}{\Delta T_2} = \frac{T_{avg1} - T_s}{T_{avg2} - T_s} \]
- Substituting the values:
\[ \frac{10}{8} = \frac{55 - T_s}{46 - T_s} \]
- Simplifying the left side:
\[ \frac{5}{4} = \frac{55 - T_s}{46 - T_s} \]

5. **Cross-multiply and solve for \(T_s\)**:
- Cross-multiplying gives:
\[ 5(46 - T_s) = 4(55 - T_s) \]
- Expanding both sides:
\[ 230 - 5T_s = 220 - 4T_s \]
- Rearranging the equation:
\[ 230 - 220 = 5T_s - 4T_s \]
\[ 10 = T_s \]

6. **Conclusion**:
- The temperature of the surrounding is:
\[ T_s = 10^\circ C \]

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

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