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the specific heat of liquid ethanol,C2H5OH(l) is 2.46j/g degree celsius and the heat of vaporization is 39.3 kj/mol the boiling point of ethanol is 78.3 c what amount of etnthalpy is required to heat 50 g of liquid ethanol form 23 c to ethanol vapor a 78.3 c?

Respuesta :

Answer: The heat required for the given process is 49.52 kJ

Explanation:

The processes involved in the given problem are:  

[tex]1.)C_2H_5OH(l)(23^oC)\rightarrow C_2H_5OH(l)(78.3^oC)\\2.)C_2H_5OH(l)(78.3^oC)\rightarrow C_2H_5OH(v)(78.3^oC)[/tex]

Pressure is taken as constant.

  • For process 1:

To calculate the amount of heat absorbed at different temperature, we use the equation:

[tex]q=m\times C_{p,m}\times (T_{2}-T_{1})[/tex]

where,

q = amount of heat absorbed = ?

[tex]C_{p,m}[/tex] = specific heat capacity of liquid ethanol = 2.46 J/g°C

m = mass of ethanol = 50 g

[tex]T_2[/tex] = final temperature  = 78.3°C

[tex]T_1[/tex] = initial temperature  = 23°C

Putting values in above equation, we get:

[tex]q_1=50\times 2.46J/g^oC\times (78.3-23)^oC\\\\q_1=6801.9J[/tex]

  • For process 2:

To calculate the amount of heat released at same temperature, we use the equation:

[tex]q=m\times L_{vap}[/tex]

where,

q = amount of heat absorbed = ?

m = mass of ethanol = 50 g

[tex]L_{vap}[/tex] = latent heat of vaporization  = 39.3 kJ/mol

Converting the latent heat of vaporization in J/g, we use the conversion factor:

Molar mass of ethanol = 46 g/mol

1 kJ = 1000 J

So, [tex](\frac{39.3kJ}{1mol})\times (\frac{1000J}{1kg})\times (\frac{1mol}{46g})=854.35J/g[/tex]

Putting values in above equation, we get:

[tex]q_2=50g\times 854.35J/g=42717.5J[/tex]

Total heat absorbed = [tex]q_1+q_2[/tex]

Total heat absorbed = [tex][6801.9+42717.5]J=49519.4J=49.52kJ[/tex]

Hence, the heat required for the given process is 49.52 kJ

dsdrajlin

The enthalpy required to heat 50 g of liquid ethanol from 23 °C to ethanol vapor at 78.3 °C is 50. kJ.

We want to heat 50 g of liquid ethanol from 23 °C to ethanol vapor at 78.3 °C. We can divide this process in 2 parts:

  1. Heating of liquid ethanol from 23 °C to 78.3 °C (boiling point).
  2. Vaporization of ethanol at 78.3 °C.

1. Heating of liquid ethanol from 23 °C to 78.3 °C

To calculate the amount of heat required for this part (Q₁), we will use the following expression.

Q₁ = c × m × ΔT

Q₁ = (2.46 J/g.°C) × 50 g × (78.3 °C - 23 °C) × (1 kJ/1000 J) = 6.8 kJ

where,

  • c is the specific heat of liquid ethanol.
  • m is the mass of ethanol.
  • ΔT is the change in the temperature.

2. Vaporization of ethanol at 78.3 °C.

To calculate the amount of heat required for this part (Q₂), we will use the following expression.

Q₂ = (m/M) × ΔH°vap

Q₂ = [50 g/(46.07g/mol] × (39.3 kJ/mol) = 43 kJ

where,

  • m is the mass of ethanol.
  • M is the molar mass of ethanol.
  • ΔH°vap is the enthalpy of vaporization of ethanol.

Total amount of heat required

The total amount of heat required (Q) is the sum of the heat required in each step.

Q = Q₁ + Q₂ = 6.8 kJ + 43 kJ = 50. kJ

Assuming the process is carried out at constant pressure, the enthalpy required for the process is 50. kJ.

The enthalpy required to heat 50 g of liquid ethanol from 23 °C to ethanol vapor at 78.3 °C is 50. kJ.

Learn more about heating curves here: https://brainly.com/question/10481356

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