Chapter 4.5, Problem 62P

1 kg of oxygen is heated from 20 to 120°C. Determine the amount of heat transfer required when this is done during a (a) constant-volume process and (b) isobaric process.

FIGURE P4–62

To determine

The amount of heat transfers to constant-volume process.

The amount of heat transfers to isobaric process.

Answer to Problem 62P

The amount of heat transfers to constant-volume process is $\underset{\_}{\text{66}\text{.7}\text{kJ}}$.

The amount of heat transfers to isobaric process is $\underset{\_}{92.7\text{kJ}}$.

Explanation of Solution

Write the general expression for the energy balance equation.

$E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$ (I)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Simplify Equation (I) and write energy balance during a constant-volume process.

$Q_{\text{in}}-W_{\text{out}}=\Delta U$ (II)

Here, the heat to be transfer into the system is $Q_{\text{in}}$, the work to be done by the system is $W_{\text{out}}$, and the change in the internal energy is $\Delta U$.

Take the oxygen as the system.

Substitute $Q_{\text{in}}$ for $Q_{\text{in}}$ and $0$ for $W_{\text{out}}$ in Equation (II).

$\begin{array}{l}{Q}_{\text{in}}-0=\Delta U\\ Q_{\text{in}}=m{c}_V\left(T_2-T_1\right)\end{array}$ (III)

Here, the mass of oxygen is $m$, the constant-volume of specific heat for oxygen at room temperature is $c_V$, the initial temperature of oxygen is $T_1$, and the final temperature of oxygen is $T_2$.

Rewrite the Equation (I) and write energy balance during a constant-pressure process.

$\begin{array}{l}{Q}_{\text{in}}-W_{\text{b,out}}=\Delta U\\ Q_{\text{in}}=W_{\text{b,out}}+\Delta U\\ Q_{\text{in}}=\Delta H\\ Q_{\text{in}}=m{c}_P\left(T_2-T_1\right)\end{array}$ (IV)

Here, the heat to be transfer into the constant-pressure process is $Q_{\text{in}}$, the boundary work to be done by the constant-pressure process is $W_{\text{b,out}}$, the change in the internal energy is $\Delta U$, the change in enthalpy is $\Delta H$, and the constant-pressure of specific heat for oxygen at room temperature is $c_P$.

Since, the quasi-equilibrium process during a constant pressure $\Delta U+W_b=\Delta H$.

Determine the change in temperature.

$\Delta T=\frac{T_1+T_2}{2}$ (V)

Here, the initial temperature oxygen is $T_1$ and final temperature oxygen is $T_2$.

Write the expression for linear interpolation method.

$Y=\left(\frac{\left(X-X_1\right)\left(Y_2-Y_1\right)}{\left(X_2-X_1\right)}\right)+Y_1$ (VI)

Here, the result of constant-pressure of specific heat is $Y$, the target temperature is $X$, the initial temperature is $X_1$, the final temperature is $X_2$, the initial constant-volume specific heat is $Y_1$, and the final constant-volume specific heat is $Y_2$.

Conclusion:

Substitute $20°\text{C}$ for $T_1$ and $120°\text{C}$ for $T_2$ in Equation (V).

$\begin{array}{c}\Delta T=\frac{20°\text{C}+120°\text{C}}{2}\\ =70°\text{C}\\ =70°\text{C}+273\text{K}\\ =\text{343}\text{K}\end{array}$

Substitute $0.918\text{kJ}/\text{kg}$ for $Y_1$, $0.928\text{kJ}/\text{kg}$ for $Y_2$, 300 K for $X_1$, 350 K for $X_2$, 343 K for $X$ in Equation (III)

$\begin{array}{c}Y=\left[\begin{array}{c}\left(\left(343\text{K}-300\text{K}\right)\left(0.928\text{kJ}/\text{kg}-0.918\text{kJ}/\text{kg}\right)/\left(350\text{K}-300\text{K}\right)\right)\\ +\left(0.918\text{kJ}/\text{kg}\right)\end{array}\right]\\ =\left(\left(43\text{K}\right)\left(0.010\text{kJ}/\text{kg}\right)/\left(50\text{K}\right)\right)+\left(0.918\text{kJ}/\text{kg}\right)\\ =0.9266\text{kJ}/\text{kg}\cdot \text{K}\\ \cong 0.927\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From above calculation the constant-pressure of specific heat is $0.927\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $0.653\text{kJ}/\text{kg}$ for $Y_1$, $0.658\text{kJ}/\text{kg}$ for $Y_2$, 300 K for $X_1$, 350 K for $X_2$, 343 K for $X$ in Equation (III)

$\begin{array}{c}Y=\left[\begin{array}{c}\left(\left(343\text{K}-300\text{K}\right)\left(0.653\text{kJ}/\text{kg}-0.658\text{kJ}/\text{kg}\right)/\left(350\text{K}-300\text{K}\right)\right)\\ +\left(0.918\text{kJ}/\text{kg}\right)\end{array}\right]\\ =\left(\left(43\text{K}\right)\left(0.005\text{kJ}/\text{kg}\right)/\left(50\text{K}\right)\right)+\left(0.653\text{kJ}/\text{kg}\right)\\ =0.6573\text{kJ}/\text{kg}\cdot \text{K}\\ \cong 0.667\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

From above calculation the constant-volume of specific heat is $0.667\text{kJ}/\text{kg}\cdot \text{K}$.

From the Table A-2b “Ideal-gas specific heats of various common gases”, obtain the value of constant-volume specific heat of air at 300 K temperature as $0.718\text{kJ}/\text{kg}°\text{C}$.

Substitute $1\text{kg}$ for $m$, $0.667\text{kJ}/\text{kg}\cdot \text{K}$ for $c_V$, $120\text{K}$ for $T_2$, and $20\text{K}$ for $T_1$ in Equation (III).

$\begin{array}{c}{Q}_{in,V=const}=\left(1\text{kg}\right)\left(0.667\text{kJ}/\text{kg}\cdot \text{K}\right)\left(120\text{K}-20\text{K}\right)\\ =\left(0.667\text{kJ}/\text{K}\right)\left(100\text{K}\right)\\ =66.7\text{kJ}\end{array}$

Thus, the amount of heat transfers to constant-volume process is $\underset{\_}{\text{66}\text{.7}\text{kJ}}$.

Substitute $1\text{kg}$ for $m$, $0.927\text{kJ}/\text{kg}\cdot \text{K}$ for $c_P$, $120\text{K}$ for $T_2$, and $20\text{K}$ for $T_1$ in Equation (IV).

$\begin{array}{c}{Q}_{in,P=const}=\left(1\text{kg}\right)\left(0.927\text{kJ}/\text{kg}\cdot \text{K}\right)\left(120\text{K}-20\text{K}\right)\\ =\left(0.927\text{kJ}/\text{K}\right)\left(100\text{K}\right)\\ =92.7\text{kJ}\end{array}$

Thus, the amount of heat transfers to isobaric process is $\underset{\_}{92.7\text{kJ}}$.