Question

Ethanol melts at -114.0 ∘C. The enthalpy of fusion is 5.02 kJ/mol and the enthalpy of vaporization is 38.6 kJ/mol. The specific heats of solid and liquid ethanol are 0.970 J/g⋅K and 2.30 J/g⋅K, respectively.

How much heat is needed to convert 39.0 g of solid ethanol at -160.0 ∘C to liquid ethanol at -35.0 ∘C ?

I am really confused on this, any help would be greatly appreciated. Thanks!

Answer 1

Given,

Specific heat of solid methanol = 0.970 J / g K

Specific heat of liquid methanol = 2.3 J / g K

Enthalpy of fusion = 5.02 kJ / mol

Enthalpy of vaporization = 38.6 kJ / mol

Melting point = -114 degree C

The heat required to convert 39.0 g of solid ethanol at -160.0 ∘C to liquid ethanol at -35.0 ∘C can be broken down into 3 parts.

1) Heat needed to raise the temperature of solid methanol from -160 C to -114 C (melting point)

2) Heat needed to melt methanol at -114 C

3) Heat needed to raise the temp. of liquid methanol from -114 C to -35 C

1) Heat needed to raise the temperature of solid methanol from -160 C to -114 C (melting point)

E1 = m Cp (solid) (-114 - (-160)) = 39 x 0.97 x 46 = 1740.2 J

2) Heat needed to melt methanol at -114 C

E2 = moles of ethanol x enthalpy of fusion

Moles of ethanol = 39 / 46.07 = 0.846 moles

=> E2 = 0.846 x 5.02 = 4.2496 kJ = 4249.6 J

3) Heat needed to raise the temp. of liquid methanol from -114 C to -35 C

E3 = m Cp (for liquid) x (-35 - (-114)) = 39 x 2.3 x 79 = 7086.3 J

Total Heat required = E1 + E2 + E3

=> Total heat Required = 1740.2 + 4249.6 + 7086.3 = 13076.1 J = 13.076 kJ