Follow your Gut?

Originally posted by smomofo
It's probably easiest to think of like this: the balloon floats in the air in the car like a ball floating on water. When the car slows, the relatively heavy air "sloshes" toward the front of the car, and the balloon moves to the back of the car.

There is only so much air in a car, and when you slam on the brakes the people in the back seat do not begin to suffocate. I think if you actually tried this experiment, I bet the effect of fluid compression would be negligible and the balloon would bounce around randomly.

Anyone want to volunteer their car and $3.00 for a balloon all in the name of science?!

Originally posted by smomofo
I'm pretty sure that all the info needed is there. Perhaps the only assumption you need to make is that if you heat more water, it is because you actually need to. I'm not sure if that made sense. Do you want my explanation of the answer now, or does anyone want to take a stab at it?

Efficiency (E) is defined as follows:

E = (what you get) / (what you pay for)

In terms of the kettle, "what you get" is energy going from the kettle to the water (Qwater), and "what you pay for" is the energy supplied to the kettle (Qkettle). If we assume this is a constant pressure heating process (open to the atmosphere, like all whistling kettles), we have:

Qwater = Cp * dT * m

where "Cp" is the constant pressure heat capacity of the water, "dT" is the temperature change, and "m" is the amount of water in the kettle. The efficiency equation is then:

E = Qwater / Qkettle = Cp * dT * m / Qkettle

If we assume the amount of heat supplied to the kettle is proportional to the amount of water in the kettle, we have:

E = Cp * dT * m / k * m = Cp * dT / k

which is a constant. This is the general case for a well designed kettle. However, 30% of the heating energy is being lost from the kettle to the surroundings. What is the nature of the energy loss?

Is heat being lost because the heating element is not totally submerged in the water? If so, the efficiency will likely be higher with 1.5 L of water in the kettle. Or is it simply heat loss to the atmosphere due to the increase in temperature of the kettle walls? If so, more water will change the temperature gradients on the kettle walls (more of the wall will be hot when the water gets hot), resulting in increased heat loss and a reduction in efficiency.

The efficiency could increase, decrease, or stay the same depending on the design of the kettle. That's why I think you need more information to answer this one.

Originally posted by PBE6
Efficiency (E) is defined as follows:

E = (what you get) / (what you pay for)

In terms of the kettle, "what you get" is energy going from the kettle to the water (Qwater), and "what you pay for" is the energy supplied to the kettle (Qkettle). If we assume this is a constant pressure heating process (open to the atmosphere, like all whistling kettles), we ...[text shortened]... sign of the kettle. That's why I think you need more information to answer this one.

This kettle problem I have actually done several times and the results are always the same: the efficiency of the kettle increases as the amount of water in the kettle is increased. My explanation is as follows:

If efficiency is E = (what you get) / (what you pay for), then I would argue that (what you pay for) = (what you get + what you waste). What you waste is the amount of thermal energy gained by the kettle itself as it warms up with the water inside of it.

Therefore: E = (what you get) / (what you get + what you waste). The amount of energy wasted remains pretty constant as the amount of water changes. Therefore, as the amount of water increases (what you get), the efficiency increases.

Originally posted by PBE6
There is only so much air in a car, and when you slam on the brakes the people in the back seat do not begin to suffocate. I think if you actually tried this experiment, I bet the effect of fluid compression would be negligible and the balloon would bounce around randomly.

Anyone want to volunteer their car and $3.00 for a balloon all in the name of science?!

I would, but I drive a convertible.

Originally posted by smomofo
This kettle problem I have actually done several times and the results are always the same: the efficiency of the kettle increases as the amount of water in the kettle is increased. My explanation is as follows:

If efficiency is E = (what you get) / (what you pay for), then I would argue that (what you pay for) = (what you get + what you waste). What y ...[text shortened]... changes. Therefore, as the amount of water increases (what you get), the efficiency increases.

I have no doubt that this can be the case, but I'm curious as to how you ran your experiment. Could you give us a brief description?

But the amount of energy that goes into the kettle is only wasted if it leaves the system and enters the surroundings, or stays in the kettle without heating the water. For example, in a well-insulated adiabatic system, the energy that enters the kettle has nowhere to go except into the water, with a small fraction staying in the kettle. In a poorly insulated system, much of the heat will be lost to the surrounding air. In fact, in a very poorly insulated system it may not even be possible to heat up the water much because all the energy is being vented to the surroundings (think about a long, tall cylindrical kettle with fins attached to the top). Since that is the case, you can't assume the kettle absorbs the same amount of energy without more details.

Originally posted by smomofo
You are driving along the road in a car with the windows rolled up and a helium balloon floating around, say, in the middle of the car, when you hit the brakes. Which way does the balloon move in the car?

We have had variations of that one in here for years, the balloon goes in the opposing direction of accelleration so slowing down is an accelleration in the opposited direction of motion so the balloon goes to the back. If the car were still and accellerated forward, the balloon would go to the front.

Originally posted by sonhouse
We have had variations of that one in here for years, the balloon goes in the opposing direction of accelleration so slowing down is an accelleration in the opposited direction of motion so the balloon goes to the back. If the car were still and accellerated forward, the balloon would go to the front.

Then what happens when you drive steadily, with the helium baloon hanging in the centre of the cars volume, and run over a bump, lifting the car from the surface of the road? What happens with the baloon?

And what happens with the helium baloon if you accidentally run over an edge and experience a free fall for a while?

How does a helum baloon behave inside a space station, filled with normal air at normal pressure?

Originally posted by FabianFnas
Then what happens when you drive steadily, with the helium baloon hanging in the centre of the cars volume, and run over a bump, lifting the car from the surface of the road? What happens with the baloon?

And what happens with the helium baloon if you accidentally run over an edge and experience a free fall for a while?

How does a helum baloon behave inside a space station, filled with normal air at normal pressure?

Inside a space station, there is no air pressure gradient so it behaves just the same as a bowling ball would, it stays where it is or follows the air flow or goes away from a push.
If a balloon is in the center of a volume, it will rise to the top of the inside ceiling but if happened while the balloon was starting to rise, hitting a bump would create a small increase in the movement upwards for a short time, like the jerk problem, a 2nd order accell.

Originally posted by PBE6
[b]I have no doubt that this can be the case, but I'm curious as to how you ran your experiment. Could you give us a brief description?

But the amount of energy that goes into the kettle is only wasted if it leaves the system and enters the surroundings, or stays in the kettle without heating the water. For example, in a well-insulated adiabatic system, the e ...[text shortened]... is the case, you can't assume the kettle absorbs the same amount of energy without more details.[

I guess I don't look at things in quite the detail some of you do. For me, the fact that after the water is poured from the kettle, the kettle is still hot to the touch is enough evidence for me to find it reasonable that not all of the energy supplied to the kettle went into the water.

The experimental procedure is roughly:
1) Measure the mass of a dry kettle
2) Place about 1 L of water in the kettle
3) Measure the mass of the kettle and water
4) Measure the temperature of the water
5) Allow kettle to heat water for 5 minutes
6) Measure temperature of water after heating
7) Measure current and voltage supplied to kettle while heating

Originally posted by PBE6
You know what? I think I might have been wrong about the balloon in the car. It might have enough momentum to push aside the air, which admittedly won't be very compressed at all (slightly, but not much). I think sonhouse had a thread all about helium balloons...I'll take a look.

I did this. I didn't let the balloon float around, but had someone hold it by it's string so it wasn't touching the ceiling. You could clearly see it go in the opposite direction of everything else when I accelerated and braked hard.

lol,
Got any pictures?! Wow, I've got to try this now myself...
Here'a nice one, though it's not too hard.
You make a movie of a ball dropping. If you play this movie backwards, will it accelerate upwards?

Originally posted by agryson
You make a movie of a ball dropping. If you play this movie backwards, will it accelerate upwards?

No.

It will start with some velocity and then decellerate until it stops.

Plane on a runway conveyor belt that goes the opposite direction of the plane, at the same speed as the plane, will it take off?

Originally posted by doodinthemood
Plane on a runway conveyor belt that goes the opposite direction of the plane, at the same speed as the plane, will it take off?

Not a hope.

Originally posted by CauselessOne
Not a hope.

That's what you say when you follow your gut 😉