Respuesta :
Answer:
the amount of heat required is 6.5kJ
Explanation:
Hi !
To solve this problem follow the steps below
1.Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
2. use thermodynamic tables to determine enthalpies in the initial and final states using temperatures and atmospheric pressure, remember that enthalpies already have energy counts by phase change
h1=Enthalpy(Ethanol;T=25C;P=101,325kPa) =258.7KJ/kg
h2=Enthalpy(Ethanol;T=78C;P=101,325kPa)=400.1 KJ/kg.
3. uses the first law of thermodynamics that states that the energy that enters a system is the same that must come out, we can infer then that heat is equal to the product of mass due to the difference in enthalpies
Q=m(h2-h1)
m=mass=46g=0.046kg
Q=0.046kg(400.1 KJ/kg.-258.7KJ/kg)=6.5kJ
the amount of heat required is 6.5kJ
Based on melting and boiling points of alcohol, the total amount of heat required is;
Q_t = 25,423.57 J/g
The missing information includes;
T_melt = –114°C
T_boil = 78°C
∆H_fus = 5.02 kJ/mol
∆H_vap = 38.56 kJ/mol
C_solid = 0.97 J/g.K
C_liquid = 2.3 J/g.K
The formula for quantity of heat is;
Q = m × c × ΔT
where;
Q = quantity of heat
m = mass
c = specific heat
ΔT = change in temperature
We are given;
mass of ethanol; m = 46 g
Temperature of ethanol; T = 25°c
Now;
1) Let us find the heat needed from the ethanol in solid form to its lowest temperature.
Q_solid = m * c_solid * (T_melt - (T_melt - T)
Plugging in the relevant values gives;
Q_solid = 46 * 0.97 * (-114 - (-114 - 25))
Q_solid = 46 * 0.97 * 25
Q_solid = 1115.5 J/g
2) Let us now find the heat need to melt the solid ethanol.
Q_melt = m * ∆H_fus
We are give; ∆H_fus = 5.02 KJ/mol = 5020 J/mol
Converting to J/g if the molar mass of ethanol is 46.1 g/mol gives;
∆H_fus = 5020/46.1 J/g
∆H_fus ≈ 109 J/g
Thus;
Q_melt = 46 * 109
Q_melt = 5014 J/g
3) The heat needed to bring the ethanol in liquid form to its boiling point is;
Q_l = m * c_liquid * (T_boil - T_melt)
Plugging in the relevant values gives;
Q_l = 46 * 2.3 * (78 - (-114))
Q_l = 19,255.6 J/g
4) Let us find the heat needed to vaporize the boiling ethanol with the formula;
Q_vap = m * ∆Hvap
We are given;
∆Hvap = 38.56 kJ/mol = 38560/46.1 J/mol
∆Hvap = 0.8364 J/g
Thus;
Q_vap = 46 * 836.44
Q_vap = 38.47 J/g
Now, total heat is;
Q_t = Q_solid + Q_melt + Q_l + Q_vap
Q_t = 1115.5 + 5014 + 19255.6 + 38.47
Q_t = 25,423.57 J/g
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