Ethyl alcohol has a boiling point of 78.0 °C, a freezing point of -114 °C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of 2.43 kJ/kg-K. How much energy must be supplied to convert 1 kg of ethyl alcohol at -114 °C to vapor at 78.0 °C?

To convert 1 kg of ethyl alcohol at -114 °C to 1 kg of ethyl alcohol vapor at 78.0 °C, we need to consider the energy required for three processes: heating the solid ethyl alcohol from -114 °C to its melting point, melting the solid ethyl alcohol at its melting point, and heating the liquid ethyl alcohol from its melting point to its boiling point, and finally vaporizing the liquid ethyl alcohol at its boiling point.

First, we need to heat the solid ethyl alcohol from -114 °C to its melting point, which is 0 °C. The specific heat of ethyl alcohol is 2.43 kJ/kg-K, so the energy required for this process is:

Energy = mass × specific heat × temperature change

Energy = 1 kg × 2.43 kJ/kg-K × (0 °C - (-114 °C))

Next, we need to melt the solid ethyl alcohol at its melting point. The heat of fusion of ethyl alcohol is 109 kJ/kg, so the energy required for this process is:

Energy = mass × heat of fusion

Energy = 1 kg × 109 kJ/kg

Then, we need to heat the liquid ethyl alcohol from its melting point to its boiling point, which is 78.0 °C. The specific heat of ethyl alcohol is 2.43 kJ/kg-K, so the energy required for this process is:

Energy = mass × specific heat × temperature change

Energy = 1 kg × 2.43 kJ/kg-K × (78.0 °C - 0 °C)

Finally, we need to vaporize the liquid ethyl alcohol at its boiling point. The heat of vaporization of ethyl alcohol is 879 kJ/kg, so the energy required for this process is:

Energy = mass × heat of vaporization

Energy = 1 kg × 879 kJ/kg

To find the total energy required, we add up the energies from each process:

Total Energy = Energy from heating + Energy from fusion + Energy from heating + Energy from vaporization

Total Energy = (1 kg × 2.43 kJ/kg-K × (0 °C - (-114 °C))) + (1 kg × 109 kJ/kg) + (1 kg × 2.43 kJ/kg-K × (78.0 °C - 0 °C)) + (1 kg × 879 kJ/kg)

Calculating the values, we get:

Total Energy = 279.42 kJ + 109 kJ + 189.54 kJ + 879 kJ

Total Energy = 1456.96 kJ

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