Score
Title
435
AskScience Panel of Scientists XVII
3110
Is there a limit on how long a power cord can be?
1398
I know it takes two weeks for the flu vaccine to be fully effective. I assume effectiveness is zero right before the vaccine is administered, and maximum after two weeks. But is there a graph that shows how effectiveness changes in time?
8651
Why is the Congo River so deep?
17
Why the Antarctic ice cap stays in one place and does not drift freely like an iceberg?
42
Why is the separation constant for the radial equation of the Hydrogen atom in the form of L(L+1)?
11
Why can some viruses (smallpox, polio) be virtually eradicated while others cannot (HIV, influenza)?
3
How does a coax splitter work?
20
If the capacity of a battery charging another battery drops below that of the receiving battery, will it stop transferring electricity since the electrons will no longer prefer to leave the lower energy "state" of the drained battery?
3
Is there a link between a metal being conductive from an electric standpoint and it being magnetic? If so, what is the cause?
13
If our bodies are conductive, can holding a battery between two fingers deplete it completely?
2
What can stop and/or destroy a black hole?
40
How/why are so many mathematical proofs and theorems contingent on the Riemann hypothesis being true?
7438
Are there any predators that hunt for sport rather than for food?
6
Are there any ant species that don’t live in colonies?
24
Do they update the voyager software?
2
How did the Russian Woodpecker receiver work?
2
What is meant by the heat death of the universe?
19
Does language affect learning and studying?
2
If the strength of an acid is based on concentration, why are acids like Sulfuric Acid always considered so dangerous compared to others?
23
When water does down the drain, why does it always go down the drain in a form of whirlpool.?
12
From what I have learnt so far, refrigerators use chlorofluorocarbons for cooling. Do these chlorofluorocarbons run out after some time? If yes how are they replenished?
0
Microwave Ovens and Wi-Fi Signals Operate at The Same Frequency (2.4GHz). What Makes Microwave Ovens More Dangerous?
20
What makes things transparent?
16
What allows certain cars and airplanes to have their own Wifi?
31
What causes the thick mist/fog that I frequently see coming off of mountains in my area?
9
What exactly happens to a person's behavior after a lobotomy?
8
How Bayes rule was used to help with aiming cannons?
6
Is there a limit to the energy density of batteries?
3
Is a Colliod a state of matter?
0
What kind of waves on the Electromagnetic Spectrum do metal detectors use to isolate only metal and no other materials?
21
Many poisonous and venomous vertebrates get their toxins from toxic arthropods that form part of their diets. Why can't they just form the toxins themselves the same ways their prey do?
192
Why are hail storms so short?
6
How do particle accelerators such as the LHC detect particle collision products?
5
How do marine mammals keep their testes cool?
68
If aliens were to look at earth through a telescope from 65 million lightyears away, would they see dinosaurs?
13
Why does fire flicker?
27
What exactly is string theory and how does it work?
10732
What exactly does the cold virus do to me to make me so weak?
8
I understand conduction and radiation as modes of heat transfer, but convection confuses me. Why does fluid moving over an object remove heat from it as opposed to adding heat due to friction?
17
What happens to the brain as you fall asleep? Are certain proteins released to induce sleep? Is it seen as a voluntary or involuntary action?
2355 Hellothere_1 I actually did a calculation just like this for my physics course last week. The result was that a spherical object orbiting the sun at the same distance as Earth eventually assumes a temperature of roughly five degrees Celsius (slightly above freezing) This is in a very simplified model though. However, I can check whether I can find or reconstruct the formula to see how close to the sun you have to get. Does anyone know how hot a marshmallow needs to be to melt? **EDIT:** Okay so the formula is Distance to sun = radius sun * (T sun)^2 /(2 * (T marshmallow)^2 ) assuming the following values: radius sun = 696342 km T (sun) = 5776K marshmallow melting point = 50 degrees C = 323K We get the result of 63 370 834 km which is roughly 160 times the radius of the sun, roughly two fifth the distance between the sun and earth, and lies between the orbits of Venus and Mercury. So good news, if you are in a spaceship near Mercury and push a marshmallow out the door it will indeed melt. If you want to see how I got to the equation, [here](https://www.reddit.com/r/askscience/comments/7engan/from_my_kid_can_you_put_a_marshmallow_on_a_stick/dq6tw8b/) is a rough outline of my approach. **SECOND EDIT:** To alleviate some confusion, my calculation assumes a spherical body that receives heat radiation on the side facing the sun and radiates heat off to all sides equally. This is only the case when the entire surface has roughly the same temperature. For a small object like a marshmallow (that possibly even rotates) that is not a bad assumption but for large bodies like the moon it doesn't quite work that way. For those objects you can regard the 5°C as a very rough approximation of the average surface temperature (take this with a grain of salt, I'd have to do some calculations to be sure) but since heat traverses from one side to the other very badly the side facing the sun will be way hotter and the side facing away from the sun will be way colder. Earth has a greenhouse effect that traps heat inside the atmosphere which is why it is way hotter than 5°C even on average. Also, regarding astronauts and spaceships, unlike the marshmallow those produce heat internally which is very hard to get rid of without an atmosphere, so without a cooling system they will overheat regardless of the sun.
3250 Koooooj Near earth? No. The amount of heat from the sun just outside Earth's atmosphere isn't that much higher than what hits the surface--the atmosphere is mostly transparent. There is some distance at which you'd be close enough; I'm not in a position to do the math, but expect it to be close than Mercury. Perhaps try making a request on /r/theydidthemath if you don't get specific numbers here. At that distance it would be difficult but possible to shield a spacecraft, allowing it to be crewed. The main reactions that go into caramelization don't depend on oxygen, so I see no reason why the marshmallow wouldn't caramelize. It wouldn't catch fire if you're doing this in the vacuum of space. One notable effect of the vacuum of space is that the marshmallow would expand *substantially* as you removed it from the atmosphere. You can look up YouTube videos of marshmallows in vacuum chambers to get an idea of this. **** Edit, since everyone feels the need to make the same comments: Space can be very hot. Thermodynamics in space are quite different from on the ground, and I didn't intend to imply that *temperature achieved* would be similar just above Earth's atmosphere as it is on the ground. In space the thermodynamics of an object are almost entirely governed by radiation. Solar radiation is about 1 kW per m^2 on the surface, or about 1.3 kW just above the atmosphere; those are the same order of magnitude, hence my opening statement. ~~Ultimately I would expect a marshmallow to be fairly similar thermodynamically to the moon; both are fairly white and fairly spherical. Temperatures on the moon are not hot enough to toast a marshmallow, though they can be very hot (as much as 100 C in sunlight). Its important to note that the moon sees extreme temperature differences because there's no effective way for heat to transfer from the sunlit side to the dark side; it's too big for conduction to be relevant. On a marshmallow scale conduction would be very relevant and would allow the entire marshmallow to be a more moderate temperature than areas of the moon see.~~ I rescind any comparison to the moon. On the scale of a marshmallow conduction dominates. The side facing the sun would be quite hot and the side facing away would be quite cold, but exactly how hot and cold depend significantly on internal conduction. Regardless, no toasting could occur as that would require near perfect black body absorption and near perfect insulation. My initial estimate of "closer than Mercury" is probably closer than you need, but I stand by my claim that it's not hot enough near earth.
460 HootyToot Brief GoogleMagik has revealed the following: + an astronaut is safe in a space suit up to 248° F + a marshmallow needs to reach 235-240° F + the heat from the sun will reach 248°F around 4 million miles out + marshmallows largely consist of air, which would cause it to explode in the vacuum of space. If you can find a way around that last bit, you could theoretically roast a marshmallow on the heat of the sun safely, but just barely. You would need to be about 3 million miles away from it, which is about 90 million miles away from the earth and about 34 million miles past mercury.
60 dabman This sounds like an experiment Cody’s Lab could do in a vacuum chamber. Maybe place the marshmallow in a teeny tiny glass jar so the pressure of the marshmallow isn’t affected, then place in vacuum and subject the marshmallow to 130% of the solar light it would normally get on earth (using a magnifying glass and a circular slit perhaps). Sounds fun! I would imagine the glass jar would help conduct some of the heat, but after a few minutes, the jar and marshmallow will heat to the point of browning and melting the marshmallow.
19 ShunanaBanana Not sure if you will see this and I know it's not exactly the answer you are looking for, but you can actually toast a marshmallow with the sun here on Earth! Google home made solar ovens. It needs to be done on a bright hot day. It's a great project to do with kids! I did it with my 1st grade class and they loved it. The marshmallows don't get super toasty, but they do get melty and gooey if done properly.
17 GaydolphShitler Three factors determine the temperature of a planet (or marshmallow, I suppose) in space: * Distance from the star * Albedo * Greenhouse effect We can ignore the greenhouse effect, since marshmallows don't have an atmosphere. The equation for calculating the temperature of a non-greenhouse body orbiting the sun is: T = 279(1-a)^1/4 X (1/√d) * T is the average temperature in Kelvin * a is the albedo (which is unitless) * d is the distance from the star in AU (1 AU being the distance between the earth and the sun) * 279 comes is a constant based on the luminosity of the sun and the Stefan-Boltzmann constant which was figured out by people much smarter than I am Solving for distance from the sun gives us: d = ((279(1-a)^1/4 )/T)^2 Basically, we need the albedo and the melting temperature. We know the melting temperature of fructose is 103°C (376° Kelvin), and I'm assuming whatever else they add to a marshmallow to make it marshmallowy doesn't effect the melting temperature much. I couldn't find any information about the albedo value of a marshmallow (surprising, I know), but it will be fairly high because it is white. I'm going to assume it is similar to fresh snow, which has an albedo of roughly 80 (meaning it absorbs 20% of the light that hits it and reflects the other 80%). That gives us: d = ((279(1-.8)^1/4 )/~~254~~ *376*)^2 Edit: fat fingered a number Which, unless my algebra skills are failing me, solves out to: d = .246AU So assuming my math is right, you'd need to be a bit inside the orbit of Mercury in order to chuck a marshmallow out the door and have it cook in space. Another interesting note is the fact that marshmallows have air trapped inside of them, so throwing it out into a vacuum would cause it to expand massively and probably "pop." So you'd end up with a weird, stretched out cooked marshmallow floating in space.
2360 0 Hellothere_1 I actually did a calculation just like this for my physics course last week. The result was that a spherical object orbiting the sun at the same distance as Earth eventually assumes a temperature of roughly five degrees Celsius (slightly above freezing) This is in a very simplified model though. However, I can check whether I can find or reconstruct the formula to see how close to the sun you have to get. Does anyone know how hot a marshmallow needs to be to melt? **EDIT:** Okay so the formula is Distance to sun = radius sun * (T sun)^2 /(2 * (T marshmallow)^2 ) assuming the following values: radius sun = 696342 km T (sun) = 5776K marshmallow melting point = 50 degrees C = 323K We get the result of 63 370 834 km which is roughly 160 times the radius of the sun, roughly two fifth the distance between the sun and earth, and lies between the orbits of Venus and Mercury. So good news, if you are in a spaceship near Mercury and push a marshmallow out the door it will indeed melt. If you want to see how I got to the equation, [here](https://www.reddit.com/r/askscience/comments/7engan/from_my_kid_can_you_put_a_marshmallow_on_a_stick/dq6tw8b/) is a rough outline of my approach. **SECOND EDIT:** To alleviate some confusion, my calculation assumes a spherical body that receives heat radiation on the side facing the sun and radiates heat off to all sides equally. This is only the case when the entire surface has roughly the same temperature. For a small object like a marshmallow (that possibly even rotates) that is not a bad assumption but for large bodies like the moon it doesn't quite work that way. For those objects you can regard the 5°C as a very rough approximation of the average surface temperature (take this with a grain of salt, I'd have to do some calculations to be sure) but since heat traverses from one side to the other very badly the side facing the sun will be way hotter and the side facing away from the sun will be way colder. Earth has a greenhouse effect that traps heat inside the atmosphere which is why it is way hotter than 5°C even on average. Also, regarding astronauts and spaceships, unlike the marshmallow those produce heat internally which is very hard to get rid of without an atmosphere, so without a cooling system they will overheat regardless of the sun.
3253 0 Koooooj Near earth? No. The amount of heat from the sun just outside Earth's atmosphere isn't that much higher than what hits the surface--the atmosphere is mostly transparent. There is some distance at which you'd be close enough; I'm not in a position to do the math, but expect it to be close than Mercury. Perhaps try making a request on /r/theydidthemath if you don't get specific numbers here. At that distance it would be difficult but possible to shield a spacecraft, allowing it to be crewed. The main reactions that go into caramelization don't depend on oxygen, so I see no reason why the marshmallow wouldn't caramelize. It wouldn't catch fire if you're doing this in the vacuum of space. One notable effect of the vacuum of space is that the marshmallow would expand *substantially* as you removed it from the atmosphere. You can look up YouTube videos of marshmallows in vacuum chambers to get an idea of this. **** Edit, since everyone feels the need to make the same comments: Space can be very hot. Thermodynamics in space are quite different from on the ground, and I didn't intend to imply that *temperature achieved* would be similar just above Earth's atmosphere as it is on the ground. In space the thermodynamics of an object are almost entirely governed by radiation. Solar radiation is about 1 kW per m^2 on the surface, or about 1.3 kW just above the atmosphere; those are the same order of magnitude, hence my opening statement. ~~Ultimately I would expect a marshmallow to be fairly similar thermodynamically to the moon; both are fairly white and fairly spherical. Temperatures on the moon are not hot enough to toast a marshmallow, though they can be very hot (as much as 100 C in sunlight). Its important to note that the moon sees extreme temperature differences because there's no effective way for heat to transfer from the sunlit side to the dark side; it's too big for conduction to be relevant. On a marshmallow scale conduction would be very relevant and would allow the entire marshmallow to be a more moderate temperature than areas of the moon see.~~ I rescind any comparison to the moon. On the scale of a marshmallow conduction dominates. The side facing the sun would be quite hot and the side facing away would be quite cold, but exactly how hot and cold depend significantly on internal conduction. Regardless, no toasting could occur as that would require near perfect black body absorption and near perfect insulation. My initial estimate of "closer than Mercury" is probably closer than you need, but I stand by my claim that it's not hot enough near earth.
452 0 HootyToot Brief GoogleMagik has revealed the following: + an astronaut is safe in a space suit up to 248° F + a marshmallow needs to reach 235-240° F + the heat from the sun will reach 248°F around 4 million miles out + marshmallows largely consist of air, which would cause it to explode in the vacuum of space. If you can find a way around that last bit, you could theoretically roast a marshmallow on the heat of the sun safely, but just barely. You would need to be about 3 million miles away from it, which is about 90 million miles away from the earth and about 34 million miles past mercury.
61 0 dabman This sounds like an experiment Cody’s Lab could do in a vacuum chamber. Maybe place the marshmallow in a teeny tiny glass jar so the pressure of the marshmallow isn’t affected, then place in vacuum and subject the marshmallow to 130% of the solar light it would normally get on earth (using a magnifying glass and a circular slit perhaps). Sounds fun! I would imagine the glass jar would help conduct some of the heat, but after a few minutes, the jar and marshmallow will heat to the point of browning and melting the marshmallow.
18 0 ShunanaBanana Not sure if you will see this and I know it's not exactly the answer you are looking for, but you can actually toast a marshmallow with the sun here on Earth! Google home made solar ovens. It needs to be done on a bright hot day. It's a great project to do with kids! I did it with my 1st grade class and they loved it. The marshmallows don't get super toasty, but they do get melty and gooey if done properly.
17 0 GaydolphShitler Three factors determine the temperature of a planet (or marshmallow, I suppose) in space: * Distance from the star * Albedo * Greenhouse effect We can ignore the greenhouse effect, since marshmallows don't have an atmosphere. The equation for calculating the temperature of a non-greenhouse body orbiting the sun is: T = 279(1-a)^1/4 X (1/√d) * T is the average temperature in Kelvin * a is the albedo (which is unitless) * d is the distance from the star in AU (1 AU being the distance between the earth and the sun) * 279 comes is a constant based on the luminosity of the sun and the Stefan-Boltzmann constant which was figured out by people much smarter than I am Solving for distance from the sun gives us: d = ((279(1-a)^1/4 )/T)^2 Basically, we need the albedo and the melting temperature. We know the melting temperature of fructose is 103°C (376° Kelvin), and I'm assuming whatever else they add to a marshmallow to make it marshmallowy doesn't effect the melting temperature much. I couldn't find any information about the albedo value of a marshmallow (surprising, I know), but it will be fairly high because it is white. I'm going to assume it is similar to fresh snow, which has an albedo of roughly 80 (meaning it absorbs 20% of the light that hits it and reflects the other 80%). That gives us: d = ((279(1-.8)^1/4 )/~~254~~ *376*)^2 Edit: fat fingered a number Which, unless my algebra skills are failing me, solves out to: d = .246AU So assuming my math is right, you'd need to be a bit inside the orbit of Mercury in order to chuck a marshmallow out the door and have it cook in space. Another interesting note is the fact that marshmallows have air trapped inside of them, so throwing it out into a vacuum would cause it to expand massively and probably "pop." So you'd end up with a weird, stretched out cooked marshmallow floating in space.