First convert 32.5 g to moles. The molecular weight of C2H5OH is 24 + 16 + 6 = 46
You have 32.5/46 = 0.707 moles of ethanol.
The heat required (in kJ) is
0.707*38.56 + 32.5*50*2.3*10^-3
How much heat is required to convert 32.5 grams of ethanol at 28 C to the vapor phase at 78 C?
Ethanol (C2H5OH) boils at 75 C. Its density is 0.789 g/mL. The enthalpy of vaporization is 38.56 kJ/mol. The specific heat of liquid ethanol is 2.3 J/g-K.
2 answers
31