Question

The water in a thin steel container with a diameter of 12 cm boils. Between the bottom of the container and boiling water;
heat transfer coefficient 10 k W / m ² K
Passing into the water; heat flux 40 k W / m ²
a) Find the surface temperature of the bottom of the container.
b) Find the heat given to the water and the amount of evaporated water for 30 minutes.

Answer 1

Given that, d = 0.12 m , h = 10kW/m^2-K, heat flux (q) = 40kW/m^2

Let the surface temperature is Ts (degree)

Latent heat of vaporization for water (LH) = 2260 kJ/kg (standard value)

As there is no specific conditions mentioned so I am assuming that water is heated at atmospheric conditions. As we know that boiling point of water at atmospheric pressure is 100 degrees (373 K)

A) as we know that q = h×∆T

--> 40 = 10 × {Ts - 100}

--> Ts = 104 degrees.

B) heat given to water (Q) = q × area × time

--> Q = 40 × (π/4)*(0.12^2) × (30*60) = 814.3 kJ

As we know that this heat is utilised in evaporation of water so Q = mass × LH

--> 814.3 = m × 2260

--> m = 0.36 kg