91Ó°ÊÓ

(a) the solid phase's specific heat

In the first segment of the chart:

$$\begin{gathered} \mathrm{Δ}U=20\left[kJ\right]-0\left[kJ\right]=20\left[kJ\right] \\ \mathrm{Δ}T=-20\left[{}^{∘}\mathrm{C}\right]-\left(-40\left[{}^{∘}\mathrm{C}\right]\right)=20\left[{}^{∘}\mathrm{C}\right] \end{gathered}$$

Substituting values in the heat capacity equation:

$20\left[kJ\right]=0.500\left[kg\right]\mathrm{Δ}{C}_p\mathrm{Δ}20\left[{}^{∘}\mathrm{C}\right]$

Solving for $C_p:C_p=\frac{20\left[kJ\right]}{0.500\left[kg\left|\mathrm{Δ}20\right|°\mathrm{C}\right]}=2.0\left[\frac{kJ}{kg\mathrm{Δ}°\mathrm{C}}\right]$

There is no temperature change in the second segment; this is a phase change from solid to liquid. The melting temperature is as follows:

$T_{\text{melting}}=-20\left[{}^{∘}\mathrm{C}\right]$

Therefore, we need to use the heat of fusion equation. The heat absorbed is:

$\mathrm{Δ}U=40\left[kJ\right]-20\left[kJ\right]=20\left[kJ\right]$

In the heat of fusion equation, substituting values:

Solving for$L_f:L_f=\frac{20[\mathrm{kg}]}{0.500[\mathrm{kg}]}=40\left[\frac{\mathrm{kJ}}{\mathrm{kg}}\right]$

In the third segment, there is a temperature increment, and then we use the heat capacity equation.

$$\begin{gathered} \mathrm{Δ}U=120\left[kJ\right]-40\left[kJ\right]=80\left[kJ\right] \\ \mathrm{Δ}T=40\left[{}^{∘}\mathrm{C}\right]-\left(-20\left[{}^{∘}\mathrm{C}\right]\right)=60\left[{}^{∘}\mathrm{C}\right] \end{gathered}$$

In the heat capacity equation, substitute the following values:

$80\left[kJ\right]=0.500[\mathrm{kg}]\mathrm{Δ}{C}_p\mathrm{Δ}60\left[{}^{∘}\mathrm{C}\right]$

Solving for

$C_p=\frac{80\left[kJ\right]}{0.500\left[\mathrm{kg}\left|\mathrm{Δ}60\right|°\mathrm{C}âˆ\pounds \right.}=2.66\left[\frac{\mathrm{kJ}}{\mathrm{kg}°\mathrm{Δ}°\mathrm{C}}\right]$

There is no temperature change in the fourth segment; this is a phase change from liquid to vapour. This is the temperature at which water boils:

$T_{\text{boiling}}=40\left[{}^{∘}\mathrm{C}\right]$

The heat absorbed is:

$\mathrm{Δ}U=180\left[kJ\right]-120\left[kJ\right]=60\left[kJ\right]$

Substituting values in the heat of fusion equation:

$60\left[kJ\right]=0.500[\mathrm{kg}]\mathrm{Δ}{L}_v$

(b) specific heat of the liquid phase

$$\begin{gathered} \mathrm{Δ}U=20\left[kJ\right]-0\left[kJ\right]=20\left[kJ\right] \\ \mathrm{Δ}T=-20\left[{}^{∘}\mathrm{C}\right]-\left(-40\left[{}^{∘}\mathrm{C}\right]\right)=20\left[{}^{∘}\mathrm{C}\right] \end{gathered}$$

Substituting values in the heat capacity equation:

$20\left[kJ\right]=0.500\left[kg\right]\mathrm{Δ}{C}_p\mathrm{Δ}20\left[{}^{∘}\mathrm{C}\right]$

Solving for $:C\_\left\{p\right\}$

width="296">Cp=20[kJ]0.500[kg]Δ20°C=2.0kJkgΔ°C

There is no temperature change in the second segment; this is a phase change from solid to liquid. The melting temperature is as follows:

$L_f=\frac{20\left[kg\right]}{0.500[\mathrm{kg}]}=40\left[\frac{\mathrm{kJ}}{\mathrm{kg}}\right]$

There is a temperature increase in the third section, after which we employ the heat capacity equation.

$\mathrm{Δ}U=120\left[kJ\right]-40\left[kJ\right]=80\left[kJ\right]$

$\mathrm{Δ}T=40\left[{}^{∘}C\right]-\left(-20\left[{}^{∘}\mathrm{C}\right]\right)=60\left[{}^{∘}\mathrm{C}\right]$

Substituting values in the heat capacity equation:

$80\left[kJ\right]=0.500\left[kg\right]\mathrm{Δ}{C}_p\mathrm{Δ}60\left[{}^{∘}C\right]$

Solving for $C_p$:

$$\begin{gathered} C_p=\frac{80\left[kJ\right]}{0.500\left[kg\right]\mathrm{Δ}60\left[{}^{∘}\mathrm{C}\right]} \\ =2.66\left[\frac{\mathrm{kJ}}{kg\mathrm{Δ}°\mathrm{C}}\right] \end{gathered}$$

In the fourth segment, there is no temperature increment; this is a change of phase from liquid to vapor. This is the boiling temperature:

$T_{\text{boiling}}=40\left[{}^{∘}\mathrm{C}\right]$

The heat absorbed is:

$\mathrm{Δ}U=180\left[kJ\right]-120\left[kJ\right]=60\left[kJ\right]$

Substituting values in the heat of fusion equation:

$60[\mathrm{kJ}]=0.500[\mathrm{kg}]\mathrm{Δ}{L}_v$

Solving for$L_f=\frac{20[\mathrm{kg}âˆ\pounds }{0.500[\mathrm{kg}âˆ\pounds }=40\left[\frac{\mathrm{kJ}}{\mathrm{kg}}\right]$

There is a temperature increase in the third section, after which we employ the heat capacity equation.$L_v=\frac{60[\mathrm{kg}]}{0.500[\mathrm{kg}]}=120\left[\frac{\mathrm{kJ}}{\mathrm{kg}}\right]$

(c) melting and boiling temperatures,

In the first segment of the chart:

$$\begin{gathered} \mathrm{Δ}U=20\left[kJ\right]-0\left[kJ\right]=20\left[kJ\right] \\ \mathrm{Δ}T=-20\left[{}^{∘}\mathrm{C}\right]-\left(-40\left[{}^{∘}\mathrm{C}\right]\right)=20\left[{}^{∘}\mathrm{C}\right] \end{gathered}$$

Substituting values in the heat capacity equation:

$20\left[kJ\right]=0.500\left[kg\right]\mathrm{Δ}{C}_p\mathrm{Δ}20\left[{}^{∘}\mathrm{C}\right]$

Solving for $C\_\left\{p\right\}:$

width="296">Cp=20[kJ]0.500[kg]Δ20°C=2.0kJkgΔ°C

There is no temperature change in the second segment; this is a phase change from solid to liquid. The melting temperature is as follows:

$T_{\text{melting}}=-20\left[{}^{∘}\mathrm{C}\right]$

As a result, we must apply the heat of fusion equation. The amount of heat absorbed is:$\mathrm{Δ}U=40\left[kJ\right]-20\left[kJ\right]=20\left[kJ\right]$

In the heat of fusion equation, substituting values

$20\left[kJ\right]=0.500\left[kg\right]\mathrm{Δ}{L}_f$

Solving for $L\_\left\{f\right\}:$

$L_f=\frac{20[\mathrm{kg}âˆ\pounds }{0.500[\mathrm{kg}âˆ\pounds }=40\left[\frac{\mathrm{kJ}}{\mathrm{kg}}\right]$

In the third segment, there is a temperature increment, and then we use the heat capacity equation.

$$\begin{gathered} \mathrm{Δ}U=120\left[kJ\right]-40\left[kJ\right]=80\left[kJ\right] \\ \mathrm{Δ}T=40\left[{}^{∘}\mathrm{C}\right]-\left(-20\left[{}^{∘}\mathrm{C}\right]\right)=60\left[{}^{∘}\mathrm{C}\right] \end{gathered}$$

Substituting values in the heat capacity equation:

$80\left[kJ\right]=0.500[\mathrm{kg}]\mathrm{Δ}{C}_p\mathrm{Δ}60\left[{}^{∘}\mathrm{C}\right]$

Solving for$C\_\left\{p\right\}:$

$$\begin{gathered} C_p=\frac{80\left[kJ\right]}{0.500\left[\mathrm{kg}\left|\mathrm{Δ}60\right|°\mathrm{C}\right]} \\ =2.66\left[\frac{kJ}{kg\mathrm{Δ}°\mathrm{C}}\right] \end{gathered}$$

In the fourth segment, there is no temperature increment; this is a change of phase from liquid to vapor. This is the boiling temperature:

$T_{\text{boiling}}=40\left[{}^{∘}\mathrm{C}\right]$

The heat absorbed is:

$\mathrm{Δ}U=180\left[kJ\right]-120\left[kJ\right]=60\left[kJ\right]$

Substituting values in the heat of fusion equation:

$60\left[kJ\right]=0.500[\mathrm{kg}]\mathrm{Δ}{L}_v$

Solving for$L\_\left\{v\right\}:$

$$\begin{gathered} L_v=\frac{60[\mathrm{kg}]}{0.500[\mathrm{kg}]} \\ =120\left[\frac{\mathrm{kJ}}{\mathrm{kg}}\right] \end{gathered}$$

$Solvingfor{L}_f:$

(d) heats of fusion and vaporization

In the first segment of the chart:

$$\begin{gathered} \mathrm{Δ}U=20\left[kJ\right]-0\left[kJ\right]=20\left[kJ\right] \\ \mathrm{Δ}T=-20\left[{}^{∘}\mathrm{C}\right]-\left(-40\left[{}^{∘}\mathrm{C}\right]\right)=20\left[{}^{∘}\mathrm{C}\right] \end{gathered}$$

Substituting values in the heat capacity equation:

$20\left[kJ\right]=0.500\left[kg\right]\mathrm{Δ}{C}_p\mathrm{Δ}20\left[{}^{∘}\mathrm{C}\right]$

Solving for $C_p:$

$$\begin{gathered} C_p=\frac{20\left[kJ\right]}{\left.\left.0.500[\mathrm{kg}]\mathrm{Δ}20\right|°\mathrm{C}\right]} \\ =2.0\left[\frac{\mathrm{kJ}}{\mathrm{kg}\mathrm{Δ}°\mathrm{C}}\right] \end{gathered}$$

There is no temperature change in the second segment; this is a phase change from solid to liquid. The melting temperature is as follows:

$T_{\text{melting}}=-20\left[{}^{∘}\mathrm{C}\right]$

Therefore, we need to use the heat of fusion equation. The heat absorbed is:

$\mathrm{Δ}U=40\left[kJ\right]-20\left[kJ\right]=20\left[kJ\right]$

Substituting values in the heat of fusion equation:

$20\left[kJ\right]=0.500[\mathrm{kg}]\mathrm{Δ}{L}_f$

Solving for $L\_\left\{f\right\}:$

$$\begin{gathered} L_f=\frac{20[\mathrm{kg}]}{0.500[\mathrm{kg}]} \\ =40\left[\frac{\mathrm{kJ}}{\mathrm{kg}}\right] \end{gathered}$$

In the third segment, there is a temperature increment, and then we use the heat capacity equation.

$\mathrm{Δ}U=120\left[kJ\right]-40\left[kJ\right]=80\left[kJ\right]$

$\mathrm{Δ}T=40\left[{}^{∘}\mathrm{C}\right]-\left(-20\left[{}^{∘}\mathrm{C}\right]\right)=60\left[{}^{∘}\mathrm{C}\right]$

Substituting values in the heat capacity equation:

$80\left[kJ\right]=0.500\left[kg\right]\mathrm{Δ}{C}_p\mathrm{Δ}60\left[{}^{∘}\mathrm{C}\right]$

Solving for$C\_\left\{p\right\}:$

$$\begin{gathered} C_p=\frac{80[\mathrm{kJ}]}{0.500\left[\mathrm{kg}\left|\mathrm{Δ}60\right|°\mathrm{C}\right]} \\ =2.66\left[\frac{\mathrm{kJ}}{\mathrm{kg}\mathrm{Δ}°\mathrm{C}}\right] \end{gathered}$$

In the fourth segment, there is no temperature increment; this is a change of phase from liquid to vapor. This is the boiling temperature:

$T_{\text{boiling}}=40\left[{}^{∘}\mathrm{C}\right]$

The heat absorbed is:

$\mathrm{Δ}U=180\left[kJ\right]-120\left[kJ\right]=60\left[kJ\right]$

Substituting values in the heat of fusion equation:

$60\left[kJ\right]=0.500\left[kg\right]\mathrm{Δ}{L}_v$

Solving for $L_v$:

$$\begin{gathered} L_v=\frac{60[\mathrm{kg}]}{0.500[\mathrm{kg}]} \\ =120\left[\frac{\mathrm{kJ}}{\mathrm{kg}}\right] \end{gathered}$$