The Effect of Temperature on Buoyant Forces

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SUMMARY

The discussion centers on the impact of temperature on buoyant forces, specifically regarding a balloon designed to carry four 100-kg passengers. It is established that warmer air is less dense than cooler air, leading to a reduced buoyant force when the ambient temperature rises. The correct conclusion is that the balloon's ability to carry weight diminishes, necessitating additional sandbags for proper operation. The equation FB,net = (1.2 - air heated) x volume of balloon is critical in understanding these dynamics.

PREREQUISITES
  • Understanding of buoyant force principles
  • Familiarity with the ideal gas law
  • Knowledge of density variations with temperature
  • Ability to apply physics equations related to buoyancy
NEXT STEPS
  • Research the ideal gas law and its implications on buoyancy
  • Study the relationship between temperature and air density
  • Explore practical applications of buoyant force in hot air ballooning
  • Learn about the effects of altitude on air density and buoyancy
USEFUL FOR

Physics students, educators, and professionals in fields related to aerodynamics and fluid dynamics will benefit from this discussion, particularly those interested in the principles of buoyancy and temperature effects on gas behavior.

karamsoft
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Hi, I came across this question in a physics review pamphlet:

Homework Statement



A balloon for a county fair is designed to carry four
100-kg passengers when it is expanded to its maximum
volume. The designers assumed the balloon would
operate in ordinary spring temperatures. If, on the day
of the fair, the temperature reaches a record-breaking
maximum: (air density: 1.2)
A) the balloon will not be able to achieve its maximum
volume.
B) more sandbags will be needed for proper operation of the
balloon.
C) the total weight the balloon is able to carry will be
reduced.
D) once in flight, the balloon cannot be lowered until the
ambient temperature drops.


Homework Equations



FB,net = (1.2— air heated)x volume of balloon
FG= mg

The Attempt at a Solution



The book says the answer is C because Warmer air is less dense than cooler air. Less dense air will give a smaller buoyant force and the balloon will be able to carry less weight.

However, using the equation FB,net = (1.2— air heated)x volume of balloon
it is obvious that the less dense air will result in a larger difference between the the densities (of the cold and hot air) which will ultimately result in a larger buoyant force. Hence, we will need a larger downward force to compensate (force of gravity) by adding sandbags (so answer choice B, which is the opposite of C).

Thanks,
 
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karamsoft said:
However, using the equation FB,net = (1.2— air heated)x volume of balloon
it is obvious that the less dense air will result in a larger difference between the the densities (of the cold and hot air)
How so?
The 1.2 is the density of the 'unusually warm' air, yes? So was the expected density more or less than 1.2? Roughly what density might the air in the balloon have? Is that altered by the unusually warm day?
You might find the table here useful: http://en.wikipedia.org/wiki/Density#Air
 
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haruspex said:
How so?
The 1.2 is the density of the 'unusually warm' air, yes? So was the expected density more or less than 1.2? Roughly what density might the air in the balloon have? Is that altered by the unusually warm day?
You might find the table here useful: http://en.wikipedia.org/wiki/Density#Air

Thank you haruspex for your response,

I see what you mean. I was fixated on 1.2 kg/m3 as a constant value. Now I think about it, it is the other way around (that is the air heated inside the balloon will have the same density on whatever day since the temperature given by the fire is the same, right?). So on a warm day the density of the "ambient temperature" air will be less than 1.2 and the difference between the variables (densities) will be less and so the buoyant force will decrease as a result..
Awesome thanks!
 

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