Chapter 21, Problem 21P

(a)

To determine

The quantity of energy the burner must transfer to the air in the balloon.

Answer to Problem 21P

The quantity of energy the burner must transfer to the air in the balloon is $5.66\times {10}^7\text{J}$.

Explanation of Solution

Write the expression from ideal gas law.

  $PV=nRT$                                                                                             (I)

Here, $P$ is the pressure of the gas, $V$ is the volume of the gas, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature.

Write the expression for number of moles.

  $n=\frac{m}{M}$                                                                                                  (II)

Here, $m$ is the mass and $M$ is the molar mass

Write the expression for density.

  $\rho =\frac{m}{V}$                                                                                                   (III)

Here, $\rho$ is the density of air and $V$ is the volume.

Write the expression for total force exerts on the balloon under equilibrium condition.

  $\begin{array}{c}{\displaystyle \sum F_y}=0\\ B-F_{g,\text{hot air}}-F_{g,\text{cargo}}=0\end{array}$                                                                              (IV)

Here, $B$ is the buoyant force, $F_{g,\text{hot air}}$ is the force exerted on the hot air balloon, and $F_{g,\text{cargo}}$ is the force due to gravity exerted on the cargo.

Write the expression for quantity of energy transfer to the air contains diatomic gas.

  $\begin{array}{c}Q=n{C}_p\Delta T\\ =n{C}_p\left(T_h-T_c\right)\end{array}$                                                                                      (V)

Here, $C_p$ is the specific heat capacity at constant volume, $T_h$ is the hot temperature, and $T_c$ is the temperature in cargo contains air.

Conclusion:

Substitute equation (II) and (III) in equation (I) and rearrange it.

  $\begin{array}{c}PV=\frac{m}{M}RT\\ PM=\frac{m}{V}RT\\ PM=\rho RT\end{array}$

Rewrite the above equation for hot temperature.

  $PM={\rho }_{h}R{T}_h$                                                                                           (VI)

Here, ${\rho }_h$ is the density of hot air in the balloon.

Rewrite the equation for outside cargo air temperature.

  $PM={\rho }_{c}R{T}_c$                                                                                         (VII)

Here, ${\rho }_c$ is the density of outside air of the cargo.

Compare the equation (VI) and (VII) to solve for ${\rho }_h$.

  $\begin{array}{c}{\rho }_{h}R{T}_h={\rho }_{c}R{T}_c\\ {\rho }_h=\frac{{\rho }_c{T}_c}{T_h}\end{array}$                                                                                     (VIII)

Replace ${\rho }_{c}Vg$ for $B$, ${\rho }_{h}Vg$ for $F_{g,\text{hot air}}$, and $mg$ for $F_{g,\text{cargo}}$ in equation (IV).

  ${\rho }_{c}Vg-{\rho }_{h}Vg-mg=0$

Substitute equation (VIII) in the above equation.

  $\begin{array}{c}{\rho }_{c}Vg-\frac{{\rho }_c{T}_c}{T_h}Vg-mg=0\\ {\rho }_{c}V-\frac{{\rho }_c{T}_c}{T_h}V-m=0\end{array}$

Substitute $1.25{\text{kg/m}}^3$ for ${\rho }_c$, $400{\text{m}}^3$ for $V$, $10.0°\text{C}$ for $T_c$, $200\text{kg}$ for $m$, and $20.0°\text{C}$ for $T$ in above equation to find $T_h$.

  $\begin{array}{c}\left(1.25{\text{kg/m}}^3\right)\left(400{\text{m}}^3\right)-\frac{\left(1.25{\text{kg/m}}^3\right)\left(10.0+273\text{K}\right)}{T_h}\left(400{\text{m}}^3\right)-200\text{kg}=0\\ 300\text{kg}=\left(500\text{kg}\right)\left(\frac{283\text{K}}{T_h}\right)\\ T_h=\left(\frac{500\text{kg}}{300\text{kg}}\right)\left(283\text{K}\right)\\ =472\text{K}\end{array}$

Rewrite the expression (I) for warmed temperature.

  $\begin{array}{c}PV=n_{h}R{T}_h\\ n_h=\frac{PV}{R{T}_h}\end{array}$

Substitute above equation in the equation (V).

  $\begin{array}{c}Q=n{C}_p\left(T_h-T_c\right)\\ =\frac{PV}{R{T}_h}{C}_P\left(T_h-T_c\right)\\ =\frac{PV}{R{T}_h}\frac{7}{2}R\left(T_h-T_c\right)\end{array}$

Substitute $472\text{K}$ for $T_h$, $1.013\times {10}^5{\text{N/m}}^2$ for $P$, $400{\text{m}}^3$ for $V$, and $10.0°\text{C}$ for $T_c$ in above equation to find $Q$.

  $\begin{array}{c}Q=\frac{7}{2}\frac{\left(1.013\times {10}^5{\text{N/m}}^2\right)\left(400{\text{m}}^3\right)\left[472\text{K}-\left(10.0+273\text{K}\right)\right]}{472\text{K}}\\ =\frac{7}{2}\frac{\left(1.013\times {10}^5{\text{N/m}}^2\right)\left(400{\text{m}}^3\right)\left[472\text{K}-283\text{K}\right]}{472\text{K}}\\ =5.66\times {10}^7\text{J}\end{array}$

Therefore, the quantity of energy transfer to the air in the balloon is $5.66\times {10}^7\text{J}$.

(b)

To determine

The mass of the propane.

Answer to Problem 21P

The mass of the propane is $1.12\text{kg}$.

Explanation of Solution

Write the expression from the relation between heat energy and the quantity of heat released from the propane burned.

  $Q=mH$                                                                                                  (IX)

Here, $m$ is the mass of the propane and $H$ is the heat energy.

Rearrange the above equation for mass.

  $m=\frac{Q}{H}$                                                                                                    (X)

Conclusion:

Substitute $5.66\times {10}^7\text{J}$ for $Q$ and $50.3\text{MJ/kg}$ for $H$ in equation (VI) to find $m$.

  $\begin{array}{c}m=\frac{5.66\times {10}^7\text{J}}{\left(50.3\text{MJ/kg}\right)\left(\frac{{10}^6\text{J}}{1\text{MJ}}\right)}\\ =\frac{5.66\times {10}^7\text{J}}{50.3\times {10}^6\text{J/kg}}\\ =\frac{5.66\times {10}^7\text{J}}{5.03\times {10}^6\text{J/kg}}\\ =1.12\text{kg}\end{array}$

Therefore, the mass of the propane is $1.12\text{kg}$.