Water is to be boiled at sea level in a 30-cm-diameter stainless steel pan placed on top of a 3-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.

The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h

Cricetus Cricetus · Answer 2 · 2022-02-16T18:17:27+01:00

The rate of evaporation will be "2.871 Kg/hour".

Evaporation of water:

According to the question,

Rate of heat supplied:

= 60% of 3 kW

= 1.8 kW

= 1.8 KJ/s

Vaporization of water, [tex]\Delta H = 2257 \ KJ/Kg[/tex]

Time taken will be:

= [tex]\frac{2257}{1.8}[/tex]

= [tex]1254 \ s[/tex]

= [tex]\frac{1254}{3600} \ hour[/tex]

= [tex]0.3482 \ hour[/tex]

hence,

The rate of evaporation,

= [tex]\frac{Mass \ of \ water}{Time}[/tex]

= [tex]\frac{1}{0.3482}[/tex]

= [tex]2.871 \ Kg/hour[/tex]

Thus the above answer is right.

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