The specific latent heat of vaporization for ethanol is 838 kJ/kg, which means that 838 kJ of energy is required to vaporize 1 kg of ethanol.
Now, 400 g is 0.4 kg. Multiply 838 kJ/kg with 0.4 kg and we have the answer 335.2 kJ.
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The specific latent heat of vaporization for ethanol is 838 kJ/kg, which means that 838 kJ of energy is required to vaporize 1 kg of ethanol.
Now, 400 g is 0.4 kg. Multiply 838 kJ/kg with 0.4 kg and we have the answer 335.2 kJ.