Question

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 removed from 0.651 kg of ethyl alcohol that is initially a gas at 78.0°C so that it becomes a solid at -114°C?

Answer 1

946.92 kJ

Step-by-step explanation:

This process has 3 parts:

1. The first part, where the temperature of Ethyl alcohol remains constant and it changes from gas to liquid.

2. The second part, where the temperature drops from 78°C to -114°C

3. The third parts, where the temperature remains constant and it changes from liquid to solid.

The energy lost in a phase change is:

Q = m*cl

The energy lost because of the drop in temperature is:

`Q = m c(T_2-T_1)`

cl is the heat of vaporization or heat of fusion, depending on the type of phase change. c is the specific heat.

So, the energy lost in each part is:

1.
`Q_1 = 0.651kg*879 kJ/kg =  572.23 kJ`

2.
`Q_2 = 0.651kg*2.43 kJ/kgK(78.0^oC - (-114^oC)) = 303.73 kJ`

3.
`Q_3 = 0.651kg*109kJ/kg = 70.96 kJ`

Then, the total energy removed should be:

Q = Q1 + Q2 + Q3 = 572.23 kJ + 303.73kJ + 70.96kJ = 946.92 kJ