Question

A hot-air balloon (volume \(1000~\mathrm{m^3}\)) with gross mass (envelope, basket, payload; excluding the air inside) \(100~\mathrm{kg}\) is in hydrostatic balance. External air: \(T_\text{out}=10^\circ\mathrm{C}\), \(\rho_\text{out}=1~\mathrm{kg\,m^{-3}}\). The temperature of the air inside the balloon is ________ \(^\circ\mathrm{C}\). (Round off to the nearest integer)

Hint:

For quick estimates: \(\rho_\text{in}=\rho_\text{out}-M/V\). Then use \(T \propto 1/\rho\) at the same pressure to find \(T_\text{in}\).

Answer 1

Correct Answer: 40

Step 1 (Neutral buoyancy equation).
At hovering equilibrium: Buoyancy \(=\) Weight. \[ \rho_\text{out} V g = (\rho_\text{in} V + M_\text{gross}) g \Rightarrow \rho_\text{in} = \rho_\text{out} - \frac{M_\text{gross}}{V}. \] Insert values: \(\rho_\text{in} = 1 - \dfrac{100}{1000}=0.9~\mathrm{kg\,m^{-3}}\).

Step 2 (Ideal-gas relation at equal pressure).
At essentially the same ambient pressure, \(\rho \propto 1/T\). Thus \[ T_\text{in} = T_\text{out}\,\frac{\rho_\text{out}}{\rho_\text{in}} = 283.15\,\mathrm{K}\times \frac{1}{0.9}=314.6~\mathrm{K}. \]

Step 3 (Convert to Celsius and round).
\(T_\text{in}=314.6-273.15=41.5^\circ\mathrm{C}\ \Rightarrow\ \boxed{41^\circ\mathrm{C}}\) (nearest integer).

Step 4 (Sanity check).
Inside air must be warmer (less dense) than outside; computed \(41^\circ\mathrm{C}\) is \(\approx 31^\circ\mathrm{C}\) warmer than the ambient \(10^\circ\mathrm{C}\), consistent with a \(10%\) density reduction. Final Answer:
\[ \boxed{41^\circ\mathrm{C}} \]