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

Solutions

VerifiedThe amount of energy required $\textbf{per unit mass}$ to change the state (but not the temperature) of a particular material is its heat of transformation $L$. Thus,

$Q_t = L \cdot m$

$\textbf{The heat of vaporization}$ $L_V$ is the amount of energy per unit mass that must be added to vaporize a liquid or that must be removed to condense a gas.

$\textbf{The heat of fusion}$ $L_F$ is the amount of energy per unit mass that must be added to melt a solid or that must be removed to freeze a liquid.

If heat $Q$ is absorbed by an object that has mass $m$, the object’s temperature change $T_f - T_i$ is related to $Q$ by

$Q = c \cdot m \cdot (T_f - T_i)$

Where $c$ is the specific heat of the material making up the object.

$\textrm{\textbf{Here, we have a gas ethyl alcohol, that will encounter 3 phases to become fully solid, so let's state our rules}}$

```
$\color{#c34632} \Huge \boxed{Q_{Vaporization} = mL_v}$
$\color{#c34632} \Huge \boxed{Q_{fusion} = mL_f}$
$\color{#c34632} \Huge \boxed{Q_{a} = mc \Delta T}$
```