Is it possible to drop below ambient temperature

[mrSHEiK124]

Originally posted by: DisgruntledVirus

Originally posted by: mrSHEiK124

Originally posted by: So

Originally posted by: mrSHEiK124
If I'm getting this right, a proper example would be a room at 72F/whateverC, and a CPU that was running at <72F with only a heatsink? Impossible, it can only be as cool as the surrounding air will allow it to be. It'll transfer heat to the environment until it reaches equilibrium, or as close as the thermal properties of both mediums will allow.

Basically, but assume that the CPU is off. It starts hot, but no additional power is being input.

Still, I think it would never be able to get below ambient temperature. Ignore our concepts of hot/cold, and just look at the heat we're dealing with. It would somehow have to displace its thermal energy away into the environment, which would require some active process (phase change cooling, compressor, whatever).

If I take a massive heatsink that is chilled to sub-zero temps, and use that it will temporarily cool the proc below ambient until it warmed over time and reach equilibrium.

True, but it's not achieving <-than ambient temps "passively," it's because you used something that was <-than ambient to begin with.

 

[palswim]

Originally posted by: So
The reason this comes up, is that my boss claims that his uncle (intelligent, EE, masters degree, etc...) had a dutch oven filled with water that dropped below ambient when left outside in cool but above freezing temps. I think his uncle is missing something.

Ask the boss to ask the uncle what he meant by "Dutch Oven" - they may have just misunderstood each other.

 

[Jabbernyx]

Nope. Passive HSFs have a degC/W rating and that is not negative.
TECs are a different matter.

 

[dullard]

Originally posted by: So
Yeah, I'm betting on a mistake somewhere in here for a real world explanation, but I'm interested in the hypothetical. Is it at all possible? I say no, he says maybe given IR radiation from a black body (i.e. warm water in a black dutch oven).

Radiation heat transfer isn't very large at cool temperatures. Since you mention outside and cool in the same thread, I assume you mean the ambient is 50°F. I'll also assume the oven has a 0.5 m^2 surface area. Thus, if the oven was a perfect black body (emissivity of 1.0), it would radiate up to (5.67*10^-8 W/m^2/K^4 * 0.5 m^2 * 1.0 * (283 K)^4 = 157 W. Or, it could radiate the same amount of heat as 2.5 standard lightbulbs. In comparison, a human radiates roughly 550 W.

But, if it is a black body, it also will absorb every bit of radiation that approaches it. Lets consider just the sun. The sun hits the earth with about 1500 W/m^2 of radiation energy. If the oven had 0.1 m^2 facing the sun, it would absorb 150 W of energy. Basically, the sun itself cancels out perfectly what it radiates and the temperature doesn't change. Nearby objects or conduction will account for the remainder 7W of energy.

So, no, it cannot radiate and cool below ambient.

 

[Howard]

Originally posted by: So
The reason this comes up, is that my boss claims that his uncle (intelligent, EE, masters degree, etc...) had a dutch oven filled with water that dropped below ambient when left outside in cool but above freezing temps. I think his uncle is missing something.

Maybe the evaporating water cooled down the oven. Did it have a lid on?

 

[Jeff7]

Based on some example problems I did in my Heat Transfer class, yes, an object can drop below ambient temperature through radiation. The question is though, to what is it radiating? If you let it radiate to the night sky, then you can drop below ambient. If you're radiating to the walls of the room, which are likely to also at the same ambient temperature, then you won't be able to go below ambient.

If you're sticking the heatsink in a warm case, the walls of the case will transfer heat to the heatsink through radiation.

 

[So]

Originally posted by: dullard

Originally posted by: So
Yeah, I'm betting on a mistake somewhere in here for a real world explanation, but I'm interested in the hypothetical. Is it at all possible? I say no, he says maybe given IR radiation from a black body (i.e. warm water in a black dutch oven).

Radiation heat transfer isn't very large at cool temperatures. Since you mention outside and cool in the same thread, I assume you mean the ambient is 50°F. I'll also assume the oven has a 0.5 m^2 surface area. Thus, if the oven was a perfect black body (emissivity of 1.0), it would radiate up to (5.67*10^-8 W/m^2/K^4 * 0.5 m^2 * 1.0 * (283 K)^4 = 157 W. Or, it could radiate the same amount of heat as 2.5 standard lightbulbs. In comparison, a human radiates roughly 550 W.

But, if it is a black body, it also will absorb every bit of radiation that approaches it. Lets consider just the sun. The sun hits the earth with about 1500 W/m^2 of radiation energy. If the oven had 0.1 m^2 facing the sun, it would absorb 150 W of energy. Basically, the sun itself cancels out perfectly what it radiates and the temperature doesn't change. Nearby objects or conduction will account for the remainder 7W of energy.

So, no, it cannot radiate and cool below ambient.

Excellent! Okay....so, assume it's in the shade (under an overhang), no direct sunlight. Primary temperature input come from conduction and convection.

 

[dullard]

Originally posted by: So
Excellent! Okay....so, assume it's in the shade, no direct sunlight. Primary temperature changes come from conduction and convection.

The tree or building (or whatever is blocking the sun) will likely radiate back as much energy to the oven as the oven radiated towards that object. See Jeff7's post.

 

[bobsmith1492]

http://en.wikipedia.org/wiki/Evaporative_cooling

 

[So]

Originally posted by: dullard

Originally posted by: So
Excellent! Okay....so, assume it's in the shade, no direct sunlight. Primary temperature changes come from conduction and convection.

The tree or building (or whatever is blocking the sun) will likely radiate back as much energy to the oven as the oven radiated towards that object. See Jeff7's post.

Gotcha. Okay, what about Jeff7's comment about the night sky. Could, given a metal sealed pot, on wood blocks (i.e. isolated from the below ambient ground), at night, in the open, drop below ambient?

 

[spidey07]

It's from evaporation. It's how even ancient air conditioners worked - nothing more than moving water in the home.

 

[DisgruntledVirus]

Originally posted by: mrSHEiK124

Originally posted by: DisgruntledVirus

Originally posted by: mrSHEiK124

Originally posted by: So

Originally posted by: mrSHEiK124
If I'm getting this right, a proper example would be a room at 72F/whateverC, and a CPU that was running at <72F with only a heatsink? Impossible, it can only be as cool as the surrounding air will allow it to be. It'll transfer heat to the environment until it reaches equilibrium, or as close as the thermal properties of both mediums will allow.

Basically, but assume that the CPU is off. It starts hot, but no additional power is being input.

Still, I think it would never be able to get below ambient temperature. Ignore our concepts of hot/cold, and just look at the heat we're dealing with. It would somehow have to displace its thermal energy away into the environment, which would require some active process (phase change cooling, compressor, whatever).

If I take a massive heatsink that is chilled to sub-zero temps, and use that it will temporarily cool the proc below ambient until it warmed over time and reach equilibrium.

True, but it's not achieving <-than ambient temps "passively," it's because you used something that was <-than ambient to begin with.

Not entirely. If I go outside my house and leave a heatsink outside overnight, then it will be below room temp as well. Yes I need a temperature difference, but it's possible.

If I'm not allowed to do that, then evaporation would qualify. The other option is use some form of naturally moving water.

edit: I didn't read the full op until now, and the dutch oven stuff....

 

[dullard]

Originally posted by: So
Gotcha. Okay, what about Jeff7's comment about the night sky. Could, given a metal sealed pot, on wood blocks (i.e. isolated from the below ambient ground), at night, in the open, drop below ambient?

Ok, if you could shield it from radiation heat properly from the earth and the surroundings, then yes, it could lose a bit more heat to the night sky (if cloudless) than it absorbs from radiation. It won't be a lot of heat and the ambient will constantly be heating it up. The temperature drop will be small.

I still think the evaporation cooling would be greater but that isn't allowed in this discussion.

 

[ItTheCow]

Originally posted by: dullard

Originally posted by: So
Yeah, I'm betting on a mistake somewhere in here for a real world explanation, but I'm interested in the hypothetical. Is it at all possible? I say no, he says maybe given IR radiation from a black body (i.e. warm water in a black dutch oven).

Radiation heat transfer isn't very large at cool temperatures. Since you mention outside and cool in the same thread, I assume you mean the ambient is 50°F. I'll also assume the oven has a 0.5 m^2 surface area. Thus, if the oven was a perfect black body (emissivity of 1.0), it would radiate up to (5.67*10^-8 W/m^2/K^4 * 0.5 m^2 * 1.0 * (283 K)^4 = 157 W.

Wait a minute, that equation isn't quite correct. It's

Q = sigma*E*A* (T_surface^4 - T_ambient^4).

This tells you that if the object is already at ambient temperature, it won't radiate any heat at all.

 

[dullard]

Originally posted by: ItTheCow
This tells you that if the object is already at ambient temperature, it won't radiate any heat at all.

Read my second paragraph. I split your equation into two paragraphs.

 

[Evadman]

Originally posted by: So

Originally posted by: ggnl
caused by evaporation maybe?

That might be the cause in this example, but assume that no evaporation happens (phase change cooling) for the hypothetical.

Then no. The heat has to go somewhere, and the 2nd law of thermodynamics says that heat moves from warmest to coolest. The coolest is going to be ambient, so that is as low as you can go, without using some form of energy to move the heat (evaporation, phase change cooling, peltier, etc)

2nd law FTW.

 

[DrPizza]

Originally posted by: Evadman

Originally posted by: So

Originally posted by: ggnl
caused by evaporation maybe?

That might be the cause in this example, but assume that no evaporation happens (phase change cooling) for the hypothetical.

Then no. The heat has to go somewhere, and the 2nd law of thermodynamics says that heat moves from warmest to coolest. The coolest is going to be ambient, so that is as low as you can go, without using some form of energy to move the heat (evaporation, phase change cooling, peltier, etc)

2nd law FTW.

Wow, it took forever for someone to point out the 2nd law of thermodynamics.

Answer: No.

 

[Hayabusa Rider]

Originally posted by: mrSHEiK124
If I'm getting this right, a proper example would be a room at 72F/whateverC, and a CPU that was running at <72F with only a heatsink? Impossible, it can only be as cool as the surrounding air will allow it to be. It'll transfer heat to the environment until it reaches equilibrium, or as close as the thermal properties of both mediums will allow.

Wrong forum!

Gawd, I love saying that

Air moving or not, once the temperature reaches equilibrium it's thermodynamically impossible for heat to passively move. It can't happen. If it does, then something else is going on. In principle you are suggesting that it's possible for water to spontaneously change to ice at room temperature when it and the surrounding environment are at equilibrium by the water magically shedding it's heat. Now there is a non-zero probability that the molecules will rearrange by quantum mechanics to allow this, however there's also a chance by the same mechanism that the entire universe just popped into existence with all our memories coinciding by random chance. Not likely.

 

[Eli]

No, of course not.

You would never be able to cool the heatsink below ambient even without anything transferring heat into it.

Once you attach it to something, you can really forget about it.

 

[Squisher]

This video made my head hurt. intermittent absorption refrigerator

All I could think was "break'n the law, break'n the law."

 

[Bignate603]

No, it's physically impossible, as long as there is an atmosphere around it. MAYBE if you had an object in something that was a few steps away from pure vacuum the radiation would be more than the free convection. It would have to be a pretty good vacuum though.

 

[DrPizza]

Originally posted by: Hayabusa Rider

Originally posted by: mrSHEiK124
If I'm getting this right, a proper example would be a room at 72F/whateverC, and a CPU that was running at <72F with only a heatsink? Impossible, it can only be as cool as the surrounding air will allow it to be. It'll transfer heat to the environment until it reaches equilibrium, or as close as the thermal properties of both mediums will allow.

Wrong forum!

Gawd, I love saying that

Air moving or not, once the temperature reaches equilibrium it's thermodynamically impossible for heat to passively move. It can't happen. If it does, then something else is going on. In principle you are suggesting that it's possible for water to spontaneously change to ice at room temperature when it and the surrounding environment are at equilibrium by the water magically shedding it's heat. Now there is a non-zero probability that the molecules will rearrange by quantum mechanics to allow this, however there's also a chance by the same mechanism that the entire universe just popped into existence with all our memories coinciding by random chance. Not likely.

I recently saw a really neat demonstration: water at room temperature was brought to a boil by applying a pretty strong vacuum. Then, the vacuum was removed, allowing normal air pressure inside the vessel. The water turned to ice. Re: as it boiled, heat was removed from the water. Not the same type of situation the OP is referring to though.

 

[Eli]

Originally posted by: DrPizza
I recently saw a really neat demonstration: water at room temperature was brought to a boil by applying a pretty strong vacuum. Then, the vacuum was removed, allowing normal air pressure inside the vessel. The water turned to ice. Re: as it boiled, heat was removed from the water. Not the same type of situation the OP is referring to though.

That's pretty cool!