Answer:
82.86 kJ
Step-by-step 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