A liquid initally at `70^(@)C` cools to `55^(@)C` in 5 minutes and `45^(@)C` in 10 minutes. What is the temperature of the surrounding ?

Explanation

To find the temperature of the surrounding, we can 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 (surrounding temperature).

### Step-by-Step Solution:

1. **Identify Initial and Final Temperatures**:
- Initial temperature (T1) = 70°C
- Final temperature after 5 minutes (T2) = 55°C
- Final temperature after 10 minutes (T3) = 45°C

2. **Calculate Temperature Changes**:
- Change in temperature from T1 to T2:
\[ \Delta T_1 = T1 - T2 = 70°C - 55°C = 15°C \]
- Change in temperature from T2 to T3:
\[ \Delta T_2 = T2 - T3 = 55°C - 45°C = 10°C \]

3. **Calculate Average Temperatures**:
- Average temperature during the first interval (from 70°C to 55°C):
\[ \text{Average}_1 = \frac{T1 + T2}{2} = \frac{70°C + 55°C}{2} = 62.5°C \]
- Average temperature during the second interval (from 55°C to 45°C):
\[ \text{Average}_2 = \frac{T2 + T3}{2} = \frac{55°C + 45°C}{2} = 50°C \]

4. **Set Up the Equations Using Newton's Law of Cooling**:
- For the first interval (5 minutes):
\[ \frac{\Delta T_1}{\Delta t_1} = -k \cdot (\text{Average}_1 - T_0) \]
Substituting values:
\[ \frac{15°C}{5 \text{ min}} = -k \cdot (62.5°C - T_0) \quad \text{(Equation 1)} \]

- For the second interval (10 minutes):
\[ \frac{\Delta T_2}{\Delta t_2} = -k \cdot (\text{Average}_2 - T_0) \]
Substituting values:
\[ \frac{10°C}{10 \text{ min}} = -k \cdot (50°C - T_0) \quad \text{(Equation 2)} \]

5. **Simplify the Equations**:
- From Equation 1:
\[ 3 = -k \cdot (62.5 - T_0) \]
- From Equation 2:
\[ 1 = -k \cdot (50 - T_0) \]

6. **Express k in Terms of T0**:
- From Equation 1:
\[ k = -\frac{3}{62.5 - T_0} \]
- From Equation 2:
\[ k = -\frac{1}{50 - T_0} \]

7. **Set the Equations for k Equal**:
\[ -\frac{3}{62.5 - T_0} = -\frac{1}{50 - T_0} \]
Cross-multiplying gives:
\[ 3(50 - T_0) = 1(62.5 - T_0) \]
Expanding:
\[ 150 - 3T_0 = 62.5 - T_0 \]
Rearranging:
\[ 150 - 62.5 = 3T_0 - T_0 \]
\[ 87.5 = 2T_0 \]
\[ T_0 = \frac{87.5}{2} = 43.75°C \]

8. **Final Result**:
The temperature of the surrounding (T0) is approximately **43.75°C**.