Question

Ethanol (C2H5OH) melts at -114 ∘C and boils at 78 ∘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⋅K are 2.3 J/g⋅Krespectively.

How much heat is required to convert 48.0 g of ethanol at -157 ∘C to the vapor phase at 78 ∘C?

Answer 1

Answer – We are given the, mass of ethanol = 48.0 g , ti = -157 oC , tf = 78 oC

specific heat of solid = 0.97 J/g⋅K and liquid ethanol= 2.3 J/g⋅K

First we need to calculate the heat -157 oC to -114 oC

We know the formula

q1 = m*C*∆t

    = 48.0 g * 0.97 J/g⋅K * (-114-(-157)oC

    = 2002.08 J

Heat from -114 oC to -114 oC

q2= m*∆Hfus

Moles of ethanol = 48.0 g / 46.068 g.mol^-1

                                = 1.04 moles

    q2 = 1.04 moles * 5020 J/mol

         = 5230.5 J

Heat from the -114 oC to 78 oC

q3 = m*C*∆t

    = 48.0 g * 2.3 J/g⋅K * (78-(-114)oC

    = 21196.8 J

Heat from the 78 oC to 78 o C

q4 = m*∆Hfus

   = 1.04 moles * 38560 J/mol

   = 40177.1 J

So total heat q = q1 +q2 +q3+ q4

                       = 68606 J

                  = 68.6 kJ

So, 68.6 kJ heat is required to convert 48.0 g of ethanol at -157 ∘C to the vapor phase at 78 ∘C?