A floating hot air balloon with volume 1000 m³ and gross mass (excluding the air in the balloon) 100 kg is in hydrostatic balance where the external air temperature is 10°C and density is 1 kg m³. The temperature of the air inside the balloon is ......... °C. (Round off to the nearest integer.)
[Assume that the gas constant for dry air is 287 J K\(^{-1}\) kg\(^{-1}\).]
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Solution and Explanation
- \( \rho_{{ext}} \) is the density of the external air,
- \( V \) is the volume of the balloon,
- \( g \) is the acceleration due to gravity.
The weight of the air inside the balloon is given by: \[ W_{{air}} = \rho_{{balloon}} \cdot V \cdot g \] In hydrostatic equilibrium: \[ F_b = W_{{air}} + W_{{balloon}} \] Substituting for the buoyant force and the weight of the air inside the balloon: \[ \rho_{{ext}} \cdot V \cdot g = \rho_{{balloon}} \cdot V \cdot g + W_{{balloon}} \] \[ \rho_{{ext}} = \rho_{{balloon}} + \frac{W_{{balloon}}}{V} \] Now, we calculate the temperature of the air inside the balloon. Since the density of the air inside the balloon is related to its temperature, we use the ideal gas law to relate the densities: \[ \rho_{{balloon}} = \frac{P}{R \cdot T_{{balloon}}} \] Where:
- \( P \) is the pressure (which is the same for both the external air and the air inside the balloon due to hydrostatic balance),
- \( R \) is the specific gas constant for dry air,
- \( T_{{balloon}} \) is the temperature of the air inside the balloon.
Using the relationship for density and the equation for hydrostatic equilibrium, we solve for \( T_{{balloon}} \) to obtain: \[ T_{{balloon}} = 40 \, {°C to 42°C} \] Thus, the temperature of the air inside the balloon is approximately 40 to 42°C.
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