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 0.155 m2 . The water inside the pot is at 100 ∘C, and 0.370 kg are evaporated every 2.90 min .

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

Answer 1

Solution: From the question we have

A pot with a steel bottom 8.50 mm thick rests on a hot stove.

The area of the bottom of the pot is 0.155 m2

The water inside the pot is at 100 ∘C, and 0.370 kg are evaporated every 2.90 min

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

Q=mLv=(0.370)(2256x10^3)

=8.34720x10^5J H=Q/t

=8.34720x10^5/(174)

=4.797x10^3J/s H

=kA(Th-Tc)/L TH-Tc

=HL/kA

=((4.797x10^3)(8.5x10^-3))/(50.2)(0.155)

=5.24 degree C

TH=Tc+5.24=100+5.24=105.24 degree C