Question

A pot with a steel bottom 8.50 mm thick rests on a hot stove. The area of the bottom of the pot is 0.150 m2. The water inside the pot is at 100.0o C, and 0.390 kg are evaporated every 3.00 min. Find the temperature of the lower surface of the pot, which is in contact with the stove.

Answer 1

Answer:

The temperature of the lower surface of the pot is 105.5397$\overline 3$ °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;

$K = \dfrac{Q \cdot d}{A \cdot \Delta T}$

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

$50 \ W/(m\cdot K) = \dfrac{4,888 \ W \times 0.0085 \ m}{0.15 \ m^2 \times \Delta T}$

$\Delta T = \dfrac{4,888 \ W \times 0.0085 \ m}{0.15 \ m^2 \times 50 \ W/(m\cdot K)} = 5 \dfrac{1012}{1875} \ K = 5.5397 \overline 3 \ K$

ΔT = T₂ - T₁

Therefore, T₂ = T₁ + ΔT

T₂ =  373.15 K + 5.5397$\overline 3$ K = 378.6897 K

T₂ = 378.689$\overline 3$ K = (378.689$\overline 3$ - 273.15)°C = 105.5397$\overline 3$ °C

The temperature of the lower surface of the pot = T₂ = 105.5397$\overline 3$ °C.