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.

mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h

Explanation:

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

answer:

okay

explanation: