A pot with a steel bottom 1.2 cm thick rests on a hot stove. The area of the bottom of the pot is `0.150m^2`. The water inside the pot is at `100^@C and 0.440 kg` are evaporated every 5.0 minute. Find the temperature of the lower surface of the pot, which is in contact with the stove. Take `L_v = 2.256 xx (10^6) J//kg and k_(steel) = 50.2 W//m-K` .

Explanation

To solve the problem, we need to find the temperature of the lower surface of the pot that is in contact with the stove, given the heat transfer through the steel bottom of the pot and the evaporation of water.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Thickness of the pot's bottom, \( L = 1.2 \, \text{cm} = 0.012 \, \text{m} \)
- Area of the bottom of the pot, \( A = 0.150 \, \text{m}^2 \)
- Mass of water evaporated, \( m = 0.440 \, \text{kg} \)
- Latent heat of vaporization, \( L_v = 2.256 \times 10^6 \, \text{J/kg} \)
- Thermal conductivity of steel, \( k = 50.2 \, \text{W/m-K} \)
- Temperature of water, \( T_{water} = 100^\circ C = 373 \, \text{K} \)
- Time for evaporation, \( t = 5.0 \, \text{min} = 300 \, \text{s} \)

2. **Calculate the Heat Required for Evaporation:**
\[ Q = m \cdot L_v = 0.440 \, \text{kg} \cdot 2.256 \times 10^6 \, \text{J/kg} \]
\[ Q = 993,840 \, \text{J} \]

3. **Determine the Rate of Heat Transfer:**
The rate of heat transfer through the steel bottom can be expressed using Fourier's law:
\[ \frac{Q}{t} = \frac{k \cdot A}{L} \cdot (T_{lower} - T_{water}) \]
Rearranging gives:
\[ T_{lower} - T_{water} = \frac{Q \cdot L}{k \cdot A \cdot t} \]

4. **Substitute the Values:**
\[ T_{lower} - 373 = \frac{993,840 \, \text{J} \cdot 0.012 \, \text{m}}{50.2 \, \text{W/m-K} \cdot 0.150 \, \text{m}^2 \cdot 300 \, \text{s}} \]

5. **Calculate the Right Side:**
\[ = \frac{993,840 \cdot 0.012}{50.2 \cdot 0.150 \cdot 300} \]
\[ = \frac{11,926.08}{2,265} \approx 5.27 \, \text{K} \]

6. **Find \( T_{lower} \):**
\[ T_{lower} = 373 + 5.27 = 378.27 \, \text{K} \]

7. **Convert to Celsius:**
\[ T_{lower} = 378.27 - 273 = 105.27^\circ C \]

### Final Answer:
The temperature of the lower surface of the pot, which is in contact with the stove, is approximately \( 105.27^\circ C \).