Cooking on wood burning stove

Olowkow

Philosopher
Olowkow
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We burn wood in a wood burning stove for heating our house most of the time. My lovely wife often cooks my dinner on top of the stove. One day I jokingly, and to my great regret, told her that she was using up all the heat cooking, and this was why it was chilly in the house.

She asked why, of course, and requested that I tell her exactly how much heat we were losing by her cooking my dinner. I quickly realized that I was in over my head trying to explain how heat going into the food and pots and pans could not also be efficiently radiated into the house as well. Yikes!

Question is, how much (difference in) cooling of the house takes place just by placing the cold food on the table vs. putting it on the hot stove? Need I mention that she beats me at chess most of the time?
 
Olowkow said:
Question is, how much (difference in) cooling of the house takes place just by placing the cold food on the table vs. putting it on the hot stove?
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As the problem is stated, there is no difference. Any heat put into the pan/food is still in the house and they will re-radiate that heat.

However, if the problem compares cooking on the wood stove versus your regular gas or electric stove then there is a difference. The kitchen stove is no longer introducing additional heat into the house.
 
I would have thought in this instance your house is a closed system, and as such every watt of energy emitted by your stove stays in the house. It matters not if the energy is in your belly or radiated directly into the room via the stove. Now when you wash the pots in cold water you will lose some energy down the sink.
I think.
 
Olowkow said:
We burn wood in a wood burning stove for heating our house most of the time. My lovely wife often cooks my dinner on top of the stove. One day I jokingly, and to my great regret, told her that she was using up all the heat cooking, and this was why it was chilly in the house.
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Dude, just add some more wood...(or close the window)

I use wood for heat up here in the mountains of CO...It's a matter of supply and demand, too chilly? add more wood!

Is this post a joke?

p.s. glad my dog doesn't use up all the heat by cooking.....I'd freeze
 
XBoxWarrior said:
Dude, just add some more wood...(or close the window)

I use wood for heat up here in the mountains of CO...It's a matter of supply and demand, too chilly? add more wood!

Is this post a joke?

p.s. glad my dog doesn't use up all the heat by cooking.....I'd freeze
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Hey dude, it was 2 degrees here this morning. Perhaps the smiley face might be a clue?

It is acutally a question of thermonuclear physics, since my stove is critical.

Actually it is a semi serious question. As I said in the OP, I jokingly said this to my wife, but now I am wondering whether a very small fraction of energy is somehow lost to heating of the room, since it feels like we are getting something for "nothing" by cooking on the stove. I believe pc Apeman, and George (above) have captured it, but I am just wondering if there is a deeper analysis in the physics of thermodynamics.
 
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It only makes sense that some heat (energy) would be transfered to the food from the heat radiating from the stove. The catch is that the food gives off heat too.
 
Energy is energy, and it should all eventually wind up as heat inside your house, unless it gets redirected somewhere else. There are some places where energy can be redirected:

Chemical reactions: Hypothetically, chemicals could absorb some energy, so you get less heat out afterwards. This will be tiny, however, as most of the food doesn't change, and some changes (like if you accidentally burn something) should release energy, if anything.

Steam creation: As water evaporates, or is boiled off, it absorbs a lot of energy. A fair amount of water, a few grams per cubic metre of air, can stay as vapour even at 0 C. The net result is that your house is filled with slightly cooler, slightly more humid air, which is generally a good thing in winter, as the air is very dry.

If you burn more wood when cooking, the stove needs to move air faster than normal, so not as much heat can be extracted before the CO2 filled air is vented outside.

You might be pouring warm water down the drain more while cooking/cleaning, as already mentioned.

You might be heating up the food instead of the house at first, so it seems cold before cooking, when heat is being transferred to food rather than you/the house. Afterwards it would seem warmer, when the food/pots and the stove are all working to warm up the house/you.

As for getting something for nothing, burning fuel is actually very wasteful in terms of the useful work you can get from it. If instead of just transferring the heat from burning the wood into your house, you instead used it to power a generator, and used that generator to power a heat pump, you could get maybe 3x as much heat into your house (how much this would cost is another issue). This is because high temperature heat has more potential work, compared to low temperature heat.

High temperature loosely means a few hundred degrees above room temp.

If you had some sort of fuel that burned at just above room temperature, you wouldn't be able to run this generator->heat pump and get any extra heat, because the heat is already at a low temperature, so it doesn't have any potential work. You also couldn't cook with it, naturally.
 
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Dilb said:
Energy is energy, and it should all eventually wind up as heat inside your house, unless it gets redirected somewhere else. There are some places where energy can be redirected:

Chemical reactions: Hypothetically, chemicals could absorb some energy, so you get less heat out afterwards. This will be tiny, however, as most of the food doesn't change, and some changes (like if you accidentally burn something) should release energy, if anything.

Steam creation: As water evaporates, or is boiled off, it absorbs a lot of energy. A fair amount of water, a few grams per cubic metre of air, can stay as vapour even at 0 C. The net result is that your house is filled with slightly cooler, slightly more humid air, which is generally a good thing in winter, as the air is very dry.
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Thanks. Good points. This is the sort of thing I was hoping for. Of course, any difference in heating is certainly not noticeable. I was thinking in terms of some loss like fractions of BTU's.
 
Nope, not unless you take the food outside to eat it.
 
Olowkow said:
Thanks. Good points. This is the sort of thing I was hoping for. Of course, any difference in heating is certainly not noticeable. I was thinking in terms of some loss like fractions of BTU's.
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You're welcome. It's almost certainly the case that none of these are large effects.

If you had an incredibly poorly designed stove, and while cooking burned significantly more wood than normal, causing the air to flow much faster than normal, you could have large losses, so it seems like burning more wood isn't as effective.

I looked at this (an incredibly bad heat exchanger) as one of my term projects during my engineering undergrad. I've still got the graph here. Basically, I ran hot tap water through a pipe surrounded by ice water, and measured the tap water's temperature when it came out of the pipe, as a function of how fast the water was going. One day the tap water was about 50 C, another day it was about 40 C, so there are a few different lines. The bottom number is the Reynold's number, which is just a scaled velocity. As the graph should show you, if you let the water flow slow enough, it's in the pipe for a long time, and will cool down (transferring a lot of heat). As it goes faster, the water has less time to transfer heat, so the exit temperature was higher. I.e. the graph goes up.

But as I said, any decent stove won't have this problem, because it will be designed to extract a lot of the heat while the air is moving reasonably quickly.
 

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Where is the key to that graph? What do the numbers on the bottom represent?
 
If heat could be lost, the universe would be a different place.
However heat can be converted into chemical bonds- essentially the reverse process of your wood burning stove. So if you use some of the heat from the stove to produce (say) a caramelised sauce, a little of the heat that otherwise would have been radiated is now tied up in molecular bonds. When you eat the sauce, your metabolism will recover some of this and store it as ATP to power muscle.
As no system is 100% efficient, you do, therefore, lose some heat that would otherwise be radiated.
If I lived in Colorado and had a wood stove, I'd lose precisely no sleep worrying about this. You are doubly fortunate in being able to have a wood stove and having someone to cook your dinner. Be very nice to this person.
 
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A related question:

On a hot day you decide to open your refrigerator door to cool the place down. What is the result?

ETA: assume the frig is 100% within the room.
 
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Well if the fridge uses 100 watts then that's the amount of energy the room will absorb.
 
A related question:

On a hot day you decide to open your refrigerator door to cool the place down. What is the result?

ETA: assume the frig is 100% within the room.
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A very high electric bill?
 
Rob Lister said:
A related question:

On a hot day you decide to open your refrigerator door to cool the place down. What is the result?

ETA: assume the frig is 100% within the room.
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This one I think I can answer :). The OP question I'll have to think on a little more.

Your room will warm up by the heat equivalent of the electrical power the frig consumes. Refrigerators cool their insides by making their outsides hotter -- they really just move the heat around. It takes work to do this and this will end up as heat.

An air conditioner cools a room by moving the heat from inside a room outside. Check and you will find the outside part (the Outdoor Coil or Condensing Unit) gets hot.
 
Cooking the food will absorb some heat energy that will not be re-radiated back to the house, because the act of cooking alters the chemical make-up of the food in an endothermic process.

However, your bodies use the food to produce energy, a large proportion of which is radiated by you as heat.

The difference between those two values is not something I can be bothered to calculate, largely because I'm not sure how I'd go about it.
 
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