Question

Calculate the amount of heat (in kJ) required to raise the temperature of a 79.0 g sample of ethanol from 298.0 K to 385.0 K. The specific heat capacity of ethanol is 2.42 J/g degree C. The boiling point of ethanol and Delta_vap H degree are given as 78.4 degree C and 38.56 kJ/mol, respectively.

Answer 1

Answer:

82.86 kJ

Explanation:

Given that:- The boiling point of ethanol = 78.4 °C

Also, Initial temperature = 298.0 K

Final temperature = 385.0 K

Also, T(°C) = T(K) - 273

So, Initial temperature = 298.0 - 273 °C = 25 °C

Final temperature = 385.0 - 273 °C = 112 °C

The following phase changes will be happening as:

Ethanol at 25 °C in liquid state goes to its boiling point at 78.4 °C (Q₁) and then at this temperature liquid ethanol changes to gaseous ethanol (Q₂) and then gaseous ethanol from 78.4 °C goes to 112 °C (Q₃).

m is the mass of Ethanol = 79.0 g

Q₁ is the heat gained in the temperature change from 25 °C to 78.4 °C.

Thus, Q₁ = C×ΔT

Where,  

C liquid is the specific heat of the ethanol = 2.42 J/g .°C

ΔT = 78.4 - (25) = 53.4 °C  

Since this is a difference of temperature, so,  

ΔT = 53.4 °C  

Q₂ is the enthalpy of vaporization for the given mass of Ethanol .

Thus, Q₂ = m×ΔH vaporization  

ΔH vaporization is the enthalpy of vaporization = 38.56 kJ/mol

Also, 1 mole of Ethanol = 46 g  and 1 kJ = 1000 J

So,  

ΔH vaporization is the enthalpy of vaporization = 38.56 kJ/mol = 38.56 ×1000 J / 46 g = 838.26 J/g

Q₃ is the heat absorbed in the temperature change from 78.4 °C to 112 °C.

Thus, Q₃ = C×ΔT'

Where,  

ΔT = 112 - (78.4) = 33.6 °C  

Since this is a difference of temperature, so,  

ΔT = 33.6 °C  

So,  

Q = Q₁ + Q₂ + Q₃

The total heat required =  m (C×ΔT + ΔH vaporization + C×ΔT')

Applying the values as:

Total heat = 79.0 ( 2.42×53.4 + 838.26 + 2.42×33.6) J  

                =  82855.2 J

Also 1 J  = 10⁻³ kJ

So,  

Heat required = 82.86 kJ