Question

A pot with a steel bottom 8.45 mm thick rests on a hot stove. The area of the bottom of the pot is 0.160 m2 . The water inside the pot is at 100.0 ∘C, and 0.410 kg are evaporated every 3.20 min

Find the temperature of the lower surface of the pot, which is in contact with the stove.

Answer 1

here,

mass of evaporated water, m = 0.410 kg

time, t = 3.20 min = 3*60 + 20 = 200 s

suarface are of pot, A = 0.160 m^2

thickness bottom, th = 8.45 mm = 0.00845 m

Latent heat of vaporization of water, Lv = 2257*10^3 J/kg

thermal conductivity of stainless steel, K = 15 W/m*C

heat genetrated to evaporation:
He = (Lv * m)/time
He = (2257*10^3 * 0.410)/200
He = 4626.85 J or 4626.85 Watts

From Fourier's law of heat conduction takes the form :
He = k*A*del(T)/th

4626.85 = 15 * 0.160 * del(100 - T) / 0.00845

Temperature of bottom, T = 83.71 C