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
[tex]\dot{m}=\frac{\dot{Q}}{h_{fg}}[/tex]
Where Q is the net rate heat transfer and [tex]h_{fg}[/tex] 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,
[tex]\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[/tex]
But we need this values not in seconds but in hours, so
[tex]\dot{m}=0.8*10^{-3}kg/s(3600s/1hr)\\\dot{m}=2.872kg/hr[/tex]
Hence the rate of evaporation of the water is 2.872kg per hour.
