A balloon buoyant forces?

1. Jun 9, 2014

karamsoft

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

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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,

2. Jun 9, 2014

haruspex

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

3. Jun 9, 2014

karamsoft

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!