Answer:
The temperature of the lower surface of the pot is 105.5397
°C
Explanation:
The given parameters are;
The thickness of the pot with steel bottom = 8.50 mm = 0.0085 m
The area of the bottom of the pot = 0.150 m²
The temperature of the water inside the pot, T₁ = 100.0 °C = 373.15 K
The mass of the water inside the pot evaporated every 3.00 min = 0.390 kg
The latent heat of vaporization of water, L = 2,256 × 10³ J/kg
The energy transferred per 3 minutes = 2,256 × 10³ × 0.390 = 879,840 J
The heat transferred per second, Q = 879,840 J/(3.00 × 60s) = 4888 Watts
The thermal conductivity of steel, K = 50.2 W/(m·K)
Therefore, the energy transferred is given as follows;

Where;
K = The thermal conductivity = 50.2 W/(m·K)
Q = The heat transferred = 4888 W
d = The distance between two isothermal planes = 0.0085 m
A = The surface area of the heat transfer = 0.150 m²
ΔT = The temperature difference between two isothermal planes = T₂ - T₁
T₂ = The temperature of the lower surface of the pot


ΔT = T₂ - T₁
Therefore, T₂ = T₁ + ΔT
T₂ = 373.15 K + 5.5397
K = 378.6897 K
T₂ = 378.689
K = (378.689
- 273.15)°C = 105.5397
°C
The temperature of the lower surface of the pot = T₂ = 105.5397
°C.