Question

A pot of water is boiling under one atmosphere of pressure. Assume that heat enters the pot only through its bottom, which is copper and rests on a heating element. In two minutes, the mass of water boiled away is m = 0.45 kg. The radius of the pot bottom is R = 6.5 cm, and the thickness is L = 2.0 mm. What is the temperature TE of the heating element in contact with the pot?

**can you please explain in detail**

Answer 1

Thermal conductivity for copper k = 401 W/mK

Area of plate, A = π * (6.5/100)² = 0.013270 m²

Latent heat of vaporization for water L = 2 * 2270 kJ/kg = 4540 kJ/kg

Heat required to to vaporize 0.45 kg water = 0.45 * 4540 = 2043 kJ

The rate of heat transfer = 2043 / (2*60) = 17.025 kW

Heat transfer rate from heater to pot = kA(ΔT) / t

= 401* 0.013270 * (T - 100) / (1/1000)

Energy conservation yields, 17.025 * 1000 = 401* 0.013270 * (T - 100) / (1/1000)

17025 = 5.321 * (T - 100) / 0.001

17.025 = 5.321 * (T - 100)

(T - 100) = 3.199

T = 103.19 o C