Question

6. Where you are now, you want to have tea while solving questions. The water, which is found in a thin steel teapot with a diameter of 12 cm, is boiling. The heat transport coefficient between the teapot base and the boiling water is 10 kW/m2K. Since the heat flow ingrair is 40 kW/m2;

a) Find the surface temperature of the teapot base.

b) Find the heat given to the water for 30 minutes and the amount of water evaporated.

Answer 1

(a)

To = 100^oC

Write the expression for heat flux through boiling water.

q" = h(T_s -To)

T_s = To + q" /h = 100 + 40/10 =104^oC

The surface temperature of the teapot base, T_s = 104^oC or 377K.

(b)

d = 12 cm, r = 6 cm

A = 4*pi*r² = 4*3.14159*(0.06)² = 0.04524 m²

Heat given to water, Q = q" x A x t = 40 x 0.04524 x 30 x 60

Q = 3257.28 kJ

Assume the latent heat of vaporization of water is, L = 2260 kJ/kg

Q = m x L

m = Q/L = 3257.28 / 2260 = 1.44 kg

Amount of water evaporated, m = 1.44 kg