Ethanol (C_2H_5 OH) melts at -114 degree C and boils at 78 degree C. The enthalp

Question

Ethanol (C_2H_5 OH) melts at -114 degree C and boils at 78 degree C. The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heat of solid and liquid ethanol are 0.97 J/g middot K are 2.3 J/g middot K respectively. How much heat is required to convert 31.5 g of ethanol at 32 degree C to the vapor phase at 78 degree C? Express your answer using two significant figures. How much heat is required to convert 31.5 g of ethanol at - 162 degree C to the vapor phase at 78 degree C? Express your answer using two significant figures.

Explanation / Answer

Q1

heat required for m = 31.5 g of ethanol from t1 = 32 t2 = 78°C

Q = m*C*(Tf-Ti)

C = 2.3 J/gC

Q = 31.5*2.3*(78-32) = 5433.75 J

but this is liquid, we need to turn it to vapor

this is latent heat energy:

Q = n*LHvap = masS/MW*LHavp = (31.5/46)*38.56 = 26.405 kJ = 26405 J

Total Q= 5433.75 +26405 = 5697.75 J = 5697.75/1000 = 5.69 kJ

Q2

ice --> saturated ice --> sautrated loquid --> liquid --> saturated liquid --> saturated vapor

so

Q1 = m*C*(tf-Ti) = 31.5*0.97 * (-114 - -162) = 1466.64 J

Q2 = m*LH = 31.5/46 * 5.02 =3.4376 kJ = 3437.6 J

Q3 = m*C*(Tf-Ti) = 31.5*2.3*(32--114) = 10577.7 J

Q4 = previously calculated in A = 5697.75

Total Q= 1466.64 +3437.6+10577.7 +5697.75 = 21179.69 J = 21.17 kJ

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