Cooking (Thermodynamics) Problems

  • #1
Matt Poirier

Homework Statement


1.
You are cooking a meal in your kitchen and have just put a pot of frozen soup on the stove to defrost. You turn on the burner and, when you ignite the gaseous fuel (methane or propane), there is a dramatic increase in the gas's thermal energy. Where did that thermal energy come from?
Select one:
a. The fuel molecules and oxygen molecules remain unchanged during the burning process, but they convert their kinetic energy into thermal energy.
b. The fuel molecules and oxygen molecules remain unchanged during the burning process, but they convert their potential energies into thermal energy.
c. When the fuel molecules and oxygen molecules rearrange to form water and carbon dioxide molecules, they convert chemical potential energy into thermal energy.
d. When the fuel molecules and oxygen molecules rearrange to form water and carbon dioxide molecules, they convert kinetic energy into thermal energy.
2.
With gas flame touching the bottom of the pot of frozen soup, heat flows into the pot. Why does heat flow from the flame to the pot and not from the pot to the flame?
Select one:
a. Heat travels upward, so heat flows from the flame to the pot above it.
b. The flame pushes upward on the pot as it flows around the pot, so the flame does work on the pot.
c. For statistical reasons, heat flows naturally from the hotter flame to the colder pot.
d. Heat flows naturally from the hotter flame to the colder pot because of Newton's laws of motion.
3.
Although both the inside and outside surfaces of your pot are made of stainless steel, there is a thick layer of aluminum sandwiched tightly between the two stainless steel surface layers. Why does that aluminum layer help your pot cook food evenly?
Select one:
a. The aluminum layer adds thickness to the pot without adding much mass. A thicker pot bottom cooks more evenly.
b. The aluminum layer stores thermal energy so that the pot's temperature can respond faster to adjustments of the burner's flame.
c. The aluminum layer conducts heat extremely well so that all parts of the pot bottom are approximately in thermal equilibrium.
d. The aluminum layer reflects heat and prevents the pot from scorching the food inside it.
4.
As you cook the frozen soup on the burner, the soup changes phases. Three distinct periods are noticeable. During the first period, the soup is completely solid. During the second period, liquid soup appears and gradually becomes more abundant. During the third period, the soup is completely liquid. Describe the soup's temperature during these three periods, assuming for simplicity that the soup is pure water.
Select one:
a. During the first period, the solid soup's temperature is increasing. During the second period, the melting soup's temperature is approximately constant. During the third period, the liquid soup's temperature is increasing.
b. During the first period, the solid soup's temperature is increasing. During the second period, the melting soup's temperature is increasing. During the third period, the liquid soup's temperature is increasing.
c. During the first period, the solid soup's temperature is approximately constant. During the second period, the melting soup's temperature is increasing. During the third period, the liquid soup's temperature is approximately constant.
d. During the first period, the solid soup's temperature is approximately constant. During the second period, the melting soup's temperature is approximately constant. During the third period, the liquid soup's temperature is increasing.
5.
The soup that you are cooking on the burner is entirely liquid and it is warmer than room temperature. Describe the surface of the soup as the soup's temperature continues to rise.
Select one:
a. Water molecules are leaving the liquid surface for the gas and, while water molecules are also leaving the gas for the liquid surface, there is a net flow of water molecules from the liquid to the gas. That flow remains steady as the soup's temperature increases.
b. Water molecules cannot leave the liquid surface for the gas until the soup reaches its boiling temperature.
c. Water molecules are leaving the liquid surface for the gas and also leaving the gas for the liquid surface. Below the soup's boiling temperature, the two rates are equal, so there is no net flow of water molecules from the liquid to the gas.
d. Water molecules are leaving the liquid surface for the gas at an ever increasing rate. Although water molecules are also leaving the gas for the liquid surface, there is a net flow of water molecules from the liquid to the gas. That flow increases as the soup's temperature increases.
6.
The soup has begun to boil. Describe the soup's temperature as it boils, assuming for simplicity that the soup is pure water.
Select one:
a. The soup's temperature is constant as it boils because liquid water and gaseous water can coexist only at water's boiling temperature.
b. The soup's temperature increases at an accelerating pace as its liquid boils away.
c. The soup's temperature increases steadily as its liquid boils away.
d. The soup's temperature is approximately constant as it boils because nearly all of the heat that is flowing into the soup is being used to convert liquid water into gaseous water.
7.
If you hold your hand directly above the boiling soup, it will feel extremely hot and you'll have to pull your hand away to avoid being burned. Why is the heat transfer to your hand so rapid in this situation?
Select one:
a. Convection carries heat upward in the air from the pot to your hand.
b. Radiation from the pot carries heat from the pot to your hand as infrared light.
c. The air above the pot is at the boiling temperature of water, so it is much hotter than your hand and transfers heat to your skin.
d. The steam in the air above the pot condenses on your cool hand and releases its heat of evaporation into your skin.
8.
You grab the handle of the pot and remove the boiling soup from the burner. The pot's handle is made from a tube of stainless steel. Why did they make the pot's handle from stainless steel rather than aluminum?
Select one:
a. Aluminum is too soft and an aluminum handle would bend too easily.
b. Aluminum is too good a conductor of heat and an aluminum handle would become too hot to hold.
c. Aluminum is too heavy and the handle would tip over the pot.
d. Aluminum is too chemically reactive and it would burn when exposed to the stove.
9.
You are baking vegetables in the oven. You turn on the electric heating element at the bottom of the oven. That lower heating element is called the baking element. How does heat flow from the baking element to the food?
Select one:
a. Heat flows almost exclusively via radiation.
b. Heat flows primarily via convection and radiation. Conduction conveys little heat because air is a poor conductor of heat.
c. Heat flows almost exclusively via convection.
d. Heat flows via conduction, convection, and radiation. Air is a good conductor of heat.
10.
You change how you are baking the vegetables in the oven. You turn off the electric heating element at the bottom of the oven and turn on the heating element at the top of the oven. That upper heating element is called the broiler element. How does heat flow from that broiler element to the food?
Select one:
a. Heat flows almost exclusively via convection.
b. Heat flows primarily via convection and radiation. Conduction in the air is also present, but air is a poor conductor of heat.
c. Heat flows almost exclusively via radiation.
d. Heat flows via conduction, convection, and radiation. Air is a good conductor of heat.
11.
Why do the white vegetables cook almost as fast as the dark-colored vegetables when you're using the broiler element?
Select one:
a. White vegetables don't need as much heat to cook as dark vegetables.
b. White vegetables absorb heat via convection just as quickly as dark vegetables.
c. All vegetables, whether they appear white or dark, are essentially black in the infrared.
d. White vegetables reflect heat at themselves and cook quickly as a result.
12.
You take the vegetables out of the oven and cover them with aluminum foil. How does the foil help the vegetables retain thermal energy?
Select one:
a. The foil blocks radiative heat transfer from the vegetables to their surroundings. However, the vegetables can still cool via evaporation.
b. The foil generates heat that keeps the vegetables hot.
c. The foil blocks conductive heat transfer from the vegetables to their surroundings.
d. The foil blocks radiative heat transfer from the vegetables to their surroundings. It also traps water vapor so that the vegetables cannot cool via evaporation.
[/B]


Homework Equations




The Attempt at a Solution


1. C, Combustion. One molecule of methane, combined with two oxygen molecules, react to form a carbon dioxide molecule, and two water molecules usually given off as steam or water vapor during the reaction and energy.
2. C, I can't think of a physics principle that makes heat flow from hot to cold, it is certainly not A,B, or C
3. Aluminum conducts heat very well, this allows heat so spread evenly, if it weren't the pot wouldn't heat, and if it was all aluminum, the pot would heat more at the bottom.
4. A, the temperature increases until it begins to melt, and again rises when its form is constant.
5. CB, the water rises as a gas, but when coming across lower temp. water at the surface, it condenses back into liquid, it doesn't leave until it reaches boiling temp.
6. D, it is very difficult to boil water at above 100 C.
7. D, as the water condenses coming across your cooler hand, it transfers its heat to you.
8.B, Good thermal conductors make poor pot handles. Stainless steel is a relatively poor conductor of heat and by using a thin tube of that metal, they have managed to construct a handle that stays cool even when the pot itself is very hot.
9.C, the hot air rises via convention and hence transfers its heat this way to the veggies.
10. B, conduction isn't present, but I think radiation becomes more involved here.
11. C, the broiler uses radiation, but both veggies appear black in the infrared.
12. D, it reflects thermal radiation, and is impervious to water vapor, so it doesn't allow water to cool via evaporation.
 

Answers and Replies

  • #2
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5. CB, the water rises as a gas, but when coming across lower temp. water at the surface, it condenses back into liquid, it doesn't leave until it reaches boiling temp.
THE ANSWER TO 5 IS D
9.C, the hot air rises via convention and hence transfers its heat this way to the veggies.
IN MY JUDGMENT, THE ANSWER TO 9 IS B: CONVECTION AND RADIATION

10. B, conduction isn't present, but I think radiation becomes more involved here.
IN MY JUDGMENT, THE ANSWER TO 10 IS C: RADIATION
 
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  • #3
haruspex
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2. C, I can't think of a physics principle that makes heat flow from hot to cold, it Yes to those.is certainly not A,B, or C
Pick the one of those you are least sure about and give your reasoning.
5. CB, the water rises as a gas, but when coming across lower temp. water at the surface, it condenses back into liquid, it doesn't leave until it reaches boiling temp.
If you read the choices carefully you will see that they are not describing bubbles coming up through the liquid. They describe molecules at the surface leaving as gas (or not).
6. D, it is very difficult to boil water at above 100 C.
This is a tricky one. While I would say D is true, it only raises the question of why nearly all the heat is going into evaporation. One of the others gives a more fundamental reason.
7. D
8.B
Yes to those.

9.C, the hot air rises via convention and hence transfers its heat this way to the veggies.
Why C rather than B, though?
10. B, conduction isn't present, but I think radiation becomes more involved here.
Why B rather than C, though?
11. C
12. D
Yes to those.
 
  • #4
Wes Tausend
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This looks like either physics homework, or heavy duty cooking class. I think I can do a couple for you.

1) It's possible they might be looking for 'c.', here, so you could be right. But all potential energies, even chemical, are really kinetic energy, so I would go with 'd.'
2) You picked 'c.' anyway, even though you felt, "it is certainly not A,B, or C". You are correct, it is 'd.' because the "heat" vibrations of the hotter molecules are transferred to the slow, cool ones like the collisions of billiard balls or ball bearings. It even acts a bit like Newton's Cradle.

I would guess that you are studying kinetic energy, aka thermodynamics. If so that will be your key to look for answers.
Good luck with understanding your project, Matt.

Wes
 
  • #5
haruspex
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all potential energies, even chemical, are really kinetic energy,
Eh? How do you get that?
because the "heat" vibrations of the hotter molecules are transferred to the slow, cool ones like the collisions of billiard balls
Yes, but that does not follow from Newton's Laws directly. If a fast moving ball is struck at right angles by a slower one, the energy transfer will be from the slower to the faster. D is wrong.
 
  • #6
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This looks like either physics homework, or heavy duty cooking class. I think I can do a couple for you.

1) It's possible they might be looking for 'c.', here, so you could be right. But all potential energies, even chemical, are really kinetic energy, so I would go with 'd.'
2) You picked 'c.' anyway, even though you felt, "it is certainly not A,B, or C". You are correct, it is 'd.' because the "heat" vibrations of the hotter molecules are transferred to the slow, cool ones like the collisions of billiard balls or ball bearings. It even acts a bit like Newton's Cradle.

I would guess that you are studying kinetic energy, aka thermodynamics. If so that will be your key to look for answers.
Good luck with understanding your project, Matt.

Wes
I disagree with both these answers.

In 1, the chemical reactions between the fuel and oxygen involve the energetics of breaking and making chemical bonds.

In the case of 2, the student is supposed to take into account the 2nd law of thermodynamics, involving statistical mechanics to reach the most probable states.
 
  • #7
Wes Tausend
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Eh? How do you get that?

Yes, but that does not follow from Newton's Laws directly. If a fast moving ball is struck at right angles by a slower one, the energy transfer will be from the slower to the faster. D is wrong.
I'll give it a shot at explaining my reasoning and then accept whatever you say, as I am sure you are more in keeping with the modern classroom. I'm aware that we once thought of chemical energy differently, but I thought it might have become more sophisticated by now.

The bulk of the heat is transferred by the hot burned gas molecules passing over the kettle and a lesser amount of the heat is transferred by radiation. Atoms don't really bump into one another, the electrostatic shell charge prevents actual touching. But in a way, all the transfer of the hot molecules is by radiation, some of it in more intimate proximity.

So I imagine the major electrostatic molecular impact is really also transferred as would be a large, short photon exchange, similar to normal 'distant' radiation imparting a lesser kinetic energy in smaller photons... all photons transferred to a low energy atom from a high energy (hot) atom. Here is an example of motions somewhat explaining such a scenario. The first four paragraphs are most pertinent, but I agree it may be a poor way to teach the OP thermodynamics and I defer to that.

Hmm. They can't impact at perfect right angles or they wouldn't even touch, but merely have a 'close call'. The way I see it, just off a right angle necessarily ahead of the fast ball, the slower ball will gain velocity anyway and the faster lose a bit... the greater energy is still transferred to the lesser just like heat, or rather thermodynamics. In approaching speed-equilibrium, the faster cause the average to be faster whereas the slower make sure the average is not faster than that with which any started. Perfect equilibrium is never reached of course, perhaps due to Heisenberg.

Thanks for your thoughts,
Wes

I disagree with both these answers.

In 1, the chemical reactions between the fuel and oxygen involve the energetics of breaking and making chemical bonds.

In the case of 2, the student is supposed to take into account the 2nd law of thermodynamics, involving statistical mechanics to reach the most probable states.
In the case of 1), I ascribed it as a kinetic energy, and as the same thing, but I'm willing to take a few lumps. In the case of 2), you are absolutely right. Thermodynamics is the main theme and I, a clumsy oaf, missed it. Sorry.

Thanks for your thoughts,
Wes
 
  • #8
Wes Tausend
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haruspex, in #5, when you said, "If a fast moving ball is struck at right angles by a slower one, the energy transfer will be from the slower to the faster" (struck from the side), I had disagreed for the reasons given. But if a slow moving, or stationary, ball (let us say 'pool' ball) is standing (or crossing) in front of, and is struck at right angles by the fast moving ball, the fast ball could indeed bounce back and thereafter be slower of the two, or even stopped. The previous slow ball would now be moving an additive combination of the speeds, or faster than the original fast ball, if the slower moved at all. That is regarding both balls as ideal without heat lost to impact. I believe thermodynamics works that way too, sometimes in ideal fashion. Other times a photon is given off (retransmitted) resulting in radiation loss, and imperfect transfer, just like the 'pool' balls.

I'm sorry, this moving off topic, and it is my fault. I hope the OP has gotten the answers he wanted.


Thanks again,
Wes
 
  • #9
haruspex
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The bulk of the heat is transferred by the hot burned gas molecules passing over the kettle and a lesser amount of the heat is transferred by radiation. Atoms don't really bump into one another, the electrostatic shell charge prevents actual touching. But in a way, all the transfer of the hot molecules is by radiation, some of it in more intimate proximity.
None of that appears to have anything to do with the chemical potential energy. That is not kinetic. The KE is only produced when the new chemical bonds form, releasing the CPE.
 
  • #10
haruspex
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They can't impact at perfect right angles or they wouldn't even touch,
For identical spheres you may be right, haven't checked the details. So let me change the model. Imagine a long smooth pole sliding past you on ice at some speed. You kick it on the side, at a lower speed, as it goes by. The speed of the pole will increase, and that of your foot slow.
There is no rule that says energy transfer is always from the fast object to the slower one, it just tends to be that way. So the statistical aspect is vital to the explanation.
 
  • #11
Wes Tausend
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None of that appears to have anything to do with the chemical potential energy. That is not kinetic. The KE is only produced when the new chemical bonds form, releasing the CPE.
Well then that is my mistake. I simply thought that when an electron(s) from one atom shared the motions with another atom, it worked the same way... that the 'electron' energy levels were basically changed, released or absorbed photons as heat, and locked the new motion in place as a change in kinetic energy until the reaction was reversed. Must Nature be so complicated as not?

My apologies sir,
Wes
 
  • #12
haruspex
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Well then that is my mistake. I simply thought that when an electron(s) from one atom shared the motions with another atom, it worked the same way... that the 'electron' energy levels were basically changed, released or absorbed photons as heat, and locked the new motion in place as a change in kinetic energy until the reaction was reversed. Must Nature be so complicated as not?

My apologies sir,
Wes
It's much the same as electric potential energy. When hydrogen and oxygen combine, the electrons get a bit closer to the protons, on average, releasing some PE.
 
  • #13
Wes Tausend
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For identical spheres you may be right, haven't checked the details. So let me change the model. Imagine a long smooth pole sliding past you on ice at some speed. You kick it on the side, at a lower speed, as it goes by. The speed of the pole will increase, and that of your foot slow.
There is no rule that says energy transfer is always from the fast object to the slower one, it just tends to be that way. So the statistical aspect is vital to the explanation.
Yes, that is quite true that the energy levels would behave the way you state. It does seem that the objects would tend towards equilibrium after some time, the slower gaining, the faster slowing, whether as a solid that evenly disperses the heat throughout, or a hot gas that momentarily 'envelopes' (contains) a solid and cools on it's way by.

I guess part of what else I was saying was that heat conduction and heat radiation seem to be accomplished the same way, by that of electrodynamic means (photons) since no particles ever touch (outer electrons for instance) except at the most extreme levels of nuclear disruption.

If the inverse square law of electron repulsion held, the repulsion should be infinite if any two 'point' particles could touch. In addition, there has to be some small distance that requires two lightweight electrons would only initially speed away from one another at the speed of light due to their inertial mass, no faster. It must be somehow related to Planck length. Of course we can always say that the relativistic mass increase would ensure such compliance at closer, more powerful range.

Thanks,
Wes
 
  • #14
Matt Poirier
Thank you to all who replied, I submitted and have the answers for reference,
1.C
2.C
3.C
4.A
5.D
6.D
7.D
8.B
9.B
10.C
11.C
12.D
 
  • #15
Matt Poirier
THE ANSWER TO 5 IS D

IN MY JUDGMENT, THE ANSWER TO 9 IS B: CONVECTION AND RADIATION


IN MY JUDGMENT, THE ANSWER TO 10 IS C: RADIATION
Yes, sir you are correct
 

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