Calculate the energy from the sun.

[rdfloyd]

For my engineering class, my professor asked us to estimate the total energy output of the sun. The only tools we are allowed to use (for now) are:

1) A thermometer
2) A graduated cylinder
3) A container of water
4) An umbrella

We are only allowed to use information that we ask for, or that he gives us. We can use other tools, but he has to approve them first.

I have no idea of what to ask or do. Does anyone have any suggestions of where to start?

Thanks,
Rdfloyd

 

[berkeman]

For my engineering class, my professor asked us to estimate the total energy output of the sun. The only tools we are allowed to use (for now) are:

1) A thermometer
2) A graduated cylinder
3) A container of water
4) An umbrella

We are only allowed to use information that we ask for, or that he gives us. We can use other tools, but he has to approve them first.

I have no idea of what to ask or do. Does anyone have any suggestions of where to start?

Thanks,
Rdfloyd

Welcome to the PF.

What does the energy from the sun do to stuff when it hits it? How can you use this property to calculate the energy hitting the umbrella? Then how would you extend that to the total energy output of the sun? (Hint -- how far is the umbrella from the sun?)

 

[rdfloyd]

Not trying to argue, but I met with some friends and we came up with this idea:

The specific heat of water is 4.168 J/(g*K). If I take the graduate cylinder, the container, and the thermometer, I can figure out how many joules the sun "puts out" onto my container. Then, I can use the proportion:

Energy_{output on container}/SurfaceArea_container [proportional to] Energy_sun/SurfaceArea_{earth's orbit}

where SurfaceArea_{earth's orbit} is 4[pi]R^2, where R is the radius of the Sun plus the distance of the Earth to the Sun.

I would solve for Energy_sun and get my answer.

My only issue with this, is that when I calculate the joules of the Sun, it does not account for time.As for this:

What does the energy from the sun do to stuff when it hits it? How can you use this property to calculate the energy hitting the umbrella? Then how would you extend that to the total energy output of the sun? (Hint -- how far is the umbrella from the sun?)

I don't understand how the umbrella comes into play. Should I find the surface area of the umbrella? If so, why not just use the surface area of the container?

Thanks again,
rdfloyd

 

[berkeman]

Not trying to argue, but I met with some friends and we came up with this idea:

The specific heat of water is 4.168 J/(g*K). If I take the graduate cylinder, the container, and the thermometer, I can figure out how many joules the sun "puts out" onto my container. Then, I can use the proportion:

Energy_{output on container}/SurfaceArea_container [proportional to] Energy_sun/SurfaceArea_{earth's orbit}

where SurfaceArea_{earth's orbit} is 4[pi]R^2, where R is the radius of the Sun plus the distance of the Earth to the Sun.

I would solve for Energy_sun and get my answer.

My only issue with this, is that when I calculate the joules of the Sun, it does not account for time.


As for this:


I don't understand how the umbrella comes into play. Should I find the surface area of the umbrella? If so, why not just use the surface area of the container?

Thanks again,
rdfloyd

Your thinking is sound. And you are right (you should get some extra credit for that) -- you are calculating the Power output of the sun, not the energy output over some amount of time. Good.

My impression about the umbrella (assuming it is black) is that it could be used as an efficient collector of the sun (like if it were submerged in a water pan or something.

But if your graduated cylinder is a good absorber of sunlight, that works just fine too.

 

[rdfloyd]

Your thinking is sound. And you are right (you should get some extra credit for that) -- you are calculating the Power output of the sun, not the energy output over some amount of time. Good.

My impression about the umbrella (assuming it is black) is that it could be used as an efficient collector of the sun (like if it were submerged in a water pan or something.

But if your graduated cylinder is a good absorber of sunlight, that works just fine too.

The container is plastic (like tupperware), but isn't big enough to hold the entire umbrella
(which is black). I guess I could cut the umbrella up to cover the container so that it is wrapped entirely with umbrella. I assume this won't be an issue.

I want to do this as good as possible as my teacher finds joy in chewing us out about us not thinking about the slightest factors in our estimating (such as not taking into effect the ground that the container will sit on). I just want to give him the least amount of room to formulate something against our method.

Thanks,
rdfloyd

 

[berkeman]

The container is plastic (like tupperware), but isn't big enough to hold the entire umbrella
(which is black). I guess I could cut the umbrella up to cover the container so that it is wrapped entirely with umbrella. I assume this won't be an issue.

I want to do this as good as possible as my teacher finds joy in chewing us out about us not thinking about the slightest factors in our estimating (such as not taking into effect the ground that the container will sit on). I just want to give him the least amount of room to formulate something against our method.

Thanks,
rdfloyd

Would you wrap the black fabric around the outside or inside of the cotainer? Why?

 

[rdfloyd]

Not really sure now. I was thinking about wrapping it around the outside.

The main reason was to make the container more "attractive" of the sun's rays, versus the white container reflecting them away.

 

[berkeman]

Not really sure now. I was thinking about wrapping it around the outside.

The main reason was to make the container more "attractive" of the sun's rays, versus the white container reflecting them away.

If it were a clear glass or plastic container, you would wrap the black material on the inside. Why?

If it is a white plastic container, it doesn't matter what color you paint the outside or inside. It will not work well. Why not?

 

[rdfloyd]

If it was clear, the energy could "pass through". So if you covered it with the black canvas, it would allow the heat into the container.

If it was white, it would reject the heat and light, so it wouldn't matter if it was covered.

We get to ask questions today, so I will probably come back with an updated assignment.

 

[berkeman]

If it was clear, the energy could "pass through". So if you covered it with the black canvas, it would allow the heat into the container.

If it was white, it would reject the heat and light, so it wouldn't matter if it was covered.

We get to ask questions today, so I will probably come back with an updated assignment.

Mostly correct. Is glass a good thermal conductor? Is plastic?

 

[rdfloyd]

I asked about the container, and he said to just use it and not worry about it's thermal conductivity. However, he also said that I would have to calculate the efficiency (i.e. how much energy is turned into heating the water) of the container. So it seems to me that he is contradicting himself, as it would make more sense to give us containers that were better thermal conductors (glass).

So to answer your question, glass isn't that bad of a thermal conductor, but for my purposes, it should be OK. Plastic is a thermal insulator, so it wouldn't transmit much of the energy to the water, so my efficiency would be smaller, which I would have to compensate.

 

[berkeman]

I asked about the container, and he said to just use it and not worry about it's thermal conductivity. However, he also said that I would have to calculate the efficiency (i.e. how much energy is turned into heating the water) of the container. So it seems to me that he is contradicting himself, as it would make more sense to give us containers that were better thermal conductors (glass).

So to answer your question, glass isn't that bad of a thermal conductor, but for my purposes, it should be OK. Plastic is a thermal insulator, so it wouldn't transmit much of the energy to the water, so my efficiency would be smaller, which I would have to compensate.

You have good instincts on this. Let's say you have a glass container that is nice and transparent, but is not a good thermal conductor. Where would you put the black stuff?