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A hot air balloon consists of a basket hanging beneath a large envelope filled with hot air. A typical hot air balloon has a total mass of 554 kg including passengers and it’s basket, and holds 2.62×10^3 m3 Of hot air in its envelope. If the ambient air density is 1.26 kg/m3 Determine the density of hot air inside the envelope when the balloon is naturally buoyant. Neglect the volume of air just place by the basket in passengers. ——The answer will be in kilograms per meters cubed

1 Answer

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Given

Total mass, m=554 kg

Volume,


V=2.62*10^3m^3

Ambient air density,


\rho_(air)=1.26\text{ kg/m}^3

To find

Determine the density of hot air inside the envelope when the balloon is naturally buoyant.

Step-by-step explanation

We know


\begin{gathered} F=(\rho_(air)-\rho)* V* g \\ \Rightarrow mg=(\rho_(air)-\rho)* V* g \\ \Rightarrow554=(1.26-\rho)*2.62*10^3 \\ \Rightarrow\rho=1.04\text{ kg/m}^3 \end{gathered}

Conclusion

The density of hot air is


1.04\text{ kg/m}^3

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