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 109kJ `//`kg , and a specific heat of 2.43 kJ `//kg.K`. How much energy must be removed from 0.720kg of ethyl alcohol that is initially a gas at `78.0^(@)C` so that it becomes a solid at `- 114^(@)C` ?

Answer

To find out how much energy must be removed from 0.720 kg of ethyl alcohol that is initially a gas at 78.0°C so that it becomes a solid at -114°C, we can break the process down into several steps.

### Step-by-Step Solution:

1. **Identify the phases and temperature changes:**
- The ethyl alcohol starts as a gas at 78.0°C.
- It will first condense to a liquid at 78.0°C.
- Then, it will cool from 78.0°C to -114.0°C, passing through the liquid phase.
- Finally, it will freeze to become a solid at -114.0°C.

2. **Calculate the energy removed during condensation (gas to liquid):**
- The heat of vaporization (Q1) is given as 879 kJ/kg.
- The mass (m) of ethyl alcohol is 0.720 kg.
- The energy removed during condensation:
\[ Q_1 = m \times \text{heat of vaporization} = 0.720 \, \text{kg} \times 879 \, \text{kJ/kg} = 632.88 \, \text{kJ} \]

3. **Calculate the energy removed during cooling of the liquid (from 78.0°C to -114.0°C):**
- The specific heat (c) of ethyl alcohol is given as 2.43 kJ/kg·K.
- The temperature change (ΔT) is:
\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = -114.0°C - 78.0°C = -192.0°C \]
- The energy removed during cooling:
\[ Q_2 = m \times c \times \Delta T = 0.720 \, \text{kg} \times 2.43 \, \text{kJ/kg·K} \times (-192.0 \, \text{K}) = -338.66 \, \text{kJ} \]

4. **Calculate the energy removed during freezing (liquid to solid):**
- The heat of fusion (Q3) is given as 109 kJ/kg.
- The energy removed during freezing:
\[ Q_3 = m \times \text{heat of fusion} = 0.720 \, \text{kg} \times 109 \, \text{kJ/kg} = 78.48 \, \text{kJ} \]

5. **Calculate the total energy removed:**
- The total energy removed (Q_total) is the sum of the energies calculated in the previous steps:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 \]
\[ Q_{\text{total}} = 632.88 \, \text{kJ} - 338.66 \, \text{kJ} + 78.48 \, \text{kJ} = 372.70 \, \text{kJ} \]

6. **Final Result:**
- The total energy that must be removed from the ethyl alcohol to convert it from gas at 78.0°C to solid at -114.0°C is approximately **372.70 kJ**.