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. We will use the principles of heat transfer and the given data.

### Step-by-step Solution:

1. **Identify the given data:**
- Thickness of the steel bottom of the pot, \( d = 1.2 \, \text{cm} = 0.012 \, \text{m} \)
- Area of the bottom of the pot, \( A = 0.150 \, \text{m}^2 \)
- Mass of water evaporated in 5 minutes, \( m = 0.440 \, \text{kg} \)
- 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, \( T_{water} = 100^\circ C \)

2. **Calculate the heat required for evaporation:**
\[ Q = m \cdot L_v = 0.440 \, \text{kg} \cdot 2.256 \times 10^6 \, \text{J/kg} = 993840 \, \text{J} \]

3. **Determine the rate of heat transfer (power) over 5 minutes:**
\[ \text{Time} = 5 \, \text{minutes} = 300 \, \text{seconds} \]
\[ \frac{dQ}{dt} = \frac{Q}{\text{Time}} = \frac{993840 \, \text{J}}{300 \, \text{s}} = 3312.8 \, \text{W} \]

4. **Calculate the thermal resistance (R) of the steel bottom:**
\[ R = \frac{d}{k \cdot A} = \frac{0.012 \, \text{m}}{50.2 \, \text{W/m-K} \cdot 0.150 \, \text{m}^2} = \frac{0.012}{7.53} \approx 1.59 \times 10^{-3} \, \text{K/W} \]

5. **Use the heat transfer equation to find the temperature difference:**
\[ \frac{dQ}{dt} = \frac{\Delta T}{R} \]
Rearranging gives:
\[ \Delta T = R \cdot \frac{dQ}{dt} = 1.59 \times 10^{-3} \, \text{K/W} \cdot 3312.8 \, \text{W} \approx 5.28 \, \text{K} \]

6. **Calculate the temperature of the lower surface of the pot:**
\[ T_{lower} = T_{water} + \Delta T = 100^\circ C + 5.28 \, \text{K} \approx 105.28^\circ C \]

### Final Answer:
The temperature of the lower surface of the pot, which is in contact with the stove, is approximately \( 105.28^\circ C \).

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