Question

A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is $\mathbf{500}\mathbf{.}\mathbf{0}\mathbf{}{\mathit{m}}^{\mathbf{3}}$ and the surrounding air is at, what must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at $\mathbf{15}\mathbf{°}\mathit{C}$ and atmospheric pressure is $\mathbf{1}\mathbf{.}\mathbf{23}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$.

Short answer

The temperature of hot air inside the balloon is 545.27 K.

Step-by-step solution

Definition of density

The term density may be defined as the ratio of mass and volume.

determine the temperature of hot air inside the balloon:

Consider the given data as below.

The volume, $V=500.0c{m}^3=500.0\times {10}^{-6}{m}^3$

The mass, m = 290 kg

The density of air, ${\rho }_{air}=1.23\mathrm{kg}/m^3$

The air temperature, $T_{air}={15}^{o}C=15+273.15=288.15K$

The density of hot air is ${\rho }_{Hot}$.

The hot air temperature is $T_{hot}$.

The density of hot air can be calculated as

${\rho }_{hot}={\rho }_{air}-\frac{290kg}{V}$

Put all values in the above equation.

$\begin{array}{r}{\rho }_{\text{hot}}=1.23\mathrm{kg}/{\mathrm{m}}^3-\frac{290\mathrm{kg}}{500.0{\mathrm{m}}^3}\\ =0.65\mathrm{kg}/{\mathrm{m}}^3\end{array}$

And the temperature of the hot air inside the balloon is calculated as

$\begin{array}{r}\frac{T_{\text{hot}}}{T_{\text{air}}}=\frac{{\rho }_{\text{air}}}{{\rho }_{\text{hot}}}\\ T_{\text{hot}}=\frac{T_{\text{air}}\times {\rho }_{\text{air}}}{{\rho }_{\text{hot}}}\end{array}$

Substitute known values in the above equation, and you have

$\begin{array}{r}{T}_{\text{hot}}=\frac{288.15\mathrm{K}\times 1.23\mathrm{kg}/{\mathrm{m}}^3}{0.65\mathrm{kg}/{\mathrm{m}}^3}\\ =545.27\mathrm{K}\end{array}$

Hence,the temperature of hot air inside the balloon is545.27 K.