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@krzysztof-pelc-r60x7j • Apr 2, 2014
Answer is d. 0.47 kW
because:
Heat of vaporization is calculated from a temperature of 100 degrees at a 1 atm and yet you have warmed these 2 kg of water at 100 degrees .... and hence the difference -
@ishan-nohePN • Apr 2, 2014
hfg for water at 100 deg= nearly (2675-491.17)=2183.83
2*2183.83/(30*60)=2.42
No option. How come you are getting 0.47? -
@ishan-nohePN • Apr 2, 2014
334.5 KJ/Kg is latent heat of ice and not steam. -
@krzysztof-pelc-r60x7j • Apr 8, 2014
Heat of vaporization is the difference enthalpy between the points on boundary curves
h1 = 419.71 kJ / kg, h2 = 2675.72 kJ / kg
h2-h1= 2256 kJ/kg !!!!!!!!!!!!!!!!!
Q=m*dh/3600 /heat related to hours
Q=2*2256/(30*60)=2,51 kW
Answer a. is correct - first time i go after you suggest 😀)