Question

Water is to be boiled at sea level (1 atm pressure) in a 30-cm-diameter stainless steel pan placed on top of a 12-kW electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water. (Round the final answer to three decimal places.

Answer 1

Answer:

The rate of evaporation of the water is 2.872kg per hour.

Explanation:

We need to find the rate of evaporation of water.

This rate commonly express as \dot{m} and the equation is

$\dot{m}=\frac{\dot{Q}}{h_{fg}}$

Where Q is the net rate heat transfer and $h_{fg}$ is the enthalpy -in this case for vaporization.

We know that 60% of heat is transferred to water. Then only a fraction of that heat is really the 'true' rate of heat transfer,

$\dot{Q}=0.6*3kW\\\dot{Q}=10.8kW\\\dot{m}=\frac{1.8kW}{2,256.4kJ/kg}\\\dot{m}=0.8*10^{-3}kg/s$

But we need this values not in seconds but in hours, so

$\dot{m}=0.8*10^{-3}kg/s(3600s/1hr)\\\dot{m}=2.872kg/hr$

Hence the rate of evaporation of the water is 2.872kg per hour.