How can weather conditions affect the ability of hot air balloons to float?

• A. Colder air increases air density ($\rho$), which enhances the buoyant force ($F$_B) and helps hot air balloons float more easily.
• B. Windy conditions increase the buoyant force ($F$_B), allowing hot air balloons to rise faster.
• C. Higher humidity always decreases the buoyant force ($F$_B), making it easier for hot air balloons to float.
• D. Hotter air outside the balloon increases the density difference ($\Delta \rho$), making the balloon float higher.

Answer: (A)

Verified step by step guidance

1. Understand that the buoyant force \( F_B \) acting on a hot air balloon depends on the difference in air density between the outside air and the hot air inside the balloon. The buoyant force can be expressed as \( F_B = \rho_{outside} \cdot V \cdot g \), where \( \rho_{outside} \) is the density of the outside air, \( V \) is the volume of the balloon, and \( g \) is the acceleration due to gravity.
2. Recognize that colder air is denser than warmer air, so when the outside air temperature decreases, \( \rho_{outside} \) increases. This increases the buoyant force \( F_B \), making it easier for the balloon to float.
3. Consider the effect of wind: while wind itself does not directly change the buoyant force, it can affect the balloon's motion and stability. However, the problem statement suggests wind increases \( F_B \), which is not physically accurate; instead, wind affects the balloon's dynamics rather than buoyancy.
4. Analyze humidity: higher humidity means more water vapor in the air, which actually decreases air density because water vapor is less dense than dry air. Therefore, increased humidity decreases \( \rho_{outside} \), reducing the buoyant force \( F_B \) and making it harder for the balloon to float.
5. Evaluate the effect of hotter outside air: if the outside air is hotter, its density \( \rho_{outside} \) decreases, reducing the difference in density \( \Delta \rho = \rho_{outside} - \rho_{inside} \). This reduces the buoyant force \( F_B \), making it more difficult for the balloon to rise higher.