this post was submitted on 08 Jul 2026
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Science Memes

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[–] stoicmaverick@lemmy.world 16 points 8 hours ago (3 children)

Okay, who gets to be the lucky one to calculate the amount of time that thing could heat sink a pegged, modern, 120w TDP CPU before it throttles at 100C? I'll give you a sticker.

[–] resipsaloquitur@lemmy.cafe 9 points 3 hours ago

Nice try. I’m not googling “copper pegging” again.

[–] Contramuffin@lemmy.world 12 points 4 hours ago* (last edited 3 hours ago) (4 children)

Was intrigued, so made a simulation to figure it out.

TLDR: 592.2 seconds, or 9 minutes and 52.2 seconds. Very similar to the other comment - it appears temperature differentials and heat loss to the air have opposite effects on thermal throttle time and mostly cancel themselves out. For the most part, heat transfer and heat loss appear to affect the thermal throttle time less than the sheer heat mass of the block by several multiples

Assumptions:

  • Copper's heat conductivity is 400 W/m-K, and specific heat is 0.4 J/g-K, and density is 9000 kg/m^3, and these values do not change over the range of temperatures
  • Air's heat transfer coefficient is 20 W/m^2-K and does not change over the range of temperatures
  • The surrounding air does not change in temperature and remains at room temperature (25 C)
  • The input wattage is actually 120 W and not just random marketing bullshit
  • The copper block's size is 4 cm x 4 cm x 16 cm (same as other comment)
  • The temperature within the copper block differs only by the vertical axis; it is assumed that temperature does not change if you move horizontally into the block

Modeling conditions:

  • The block is sliced into 100 equally-sized slices, stacked vertically.
  • Each slice starts off with a temperature of 25 C
  • 120 W is input directly into the bottom slice
  • Heat transfer is modeled between each slice
  • Heat loss into the air is modeled for each slice (top slice has more heat loss due to more contact with the air)
  • Temperature changes are calculated per millisecond
  • Final time is calculated by the total number of milliseconds it takes for the bottom slice to reach a temperature greater than 100 C

Fun facts I found from playing around with the model:

  • According to this model, at the time that the CPU thermal throttles, the top of the block should be 85 C
  • If we assume instantaneous heat transfer, time to thermal throttle goes up to 703 seconds (11 minutes and 43 seconds). Difference is about 2 minutes.
  • If we assume no heat loss to the air, time to thermal throttle goes down to 500.0 seconds (8 minutes and 20 seconds). Difference is about 1.5 minutes.
  • The copper block should be able to prevent throttling as long as the CPU remains idle (30W for AMD CPU's). The CPU should cap out at around 82-83 C.
  • The copper block can prevent thermal throttling for a 170 W CPU for 368.1 seconds, or 6 minutes and 8.1 seconds
[–] stoicmaverick@lemmy.world 3 points 1 hour ago

Well goddamn... Ok. Go ahead and dm me your home address, phone number, social and/or tax id number, the name of the street you grew up on, the name of your favorite teacher, the IMEI number of your cellphone, a high resolution set of your fingerprints, and a list of your three greatest fears, and I'll get your sticker sent over as soon as I can.

[–] Bahnd@lemmy.world 2 points 1 hour ago

You did the monster math.

Respect.

[–] kahjtheundedicated@lemmy.world 2 points 4 hours ago

Respect for taking the time to model that. Goes to show why heat sinks look the way they do, and not just big lumps of metal lol

[–] mnemonicmonkeys@sh.itjust.works 2 points 4 hours ago (1 children)

Numerical methods is cheating! Real men use PDE's!

/s of course, though I was kinda hoping you'd use PDE's

[–] Contramuffin@lemmy.world 1 points 3 hours ago

See, I thought about doing that, but then I realized: I don't actually want to do that

[–] KSPAtlas@sopuli.xyz 26 points 7 hours ago (3 children)

Let's assume the dimensions of the copper block are 40mm40mm160mm (I'm not taking the heat spreader into account here)

That results in a volume of 256000mm3, or 256cm3

Copper (at 20C) has a density of 8.935 g/cm3, so that's roughly 2.28736kg of copper

Copper has a specific heat capacity of 384.603 J/(kg K)

Using E=cm∆t, we can figure out that it would take ≈ 70378J of energy to heat the copper block to 100C, starting at 20C

With a TDP od 120W, that means it would take 586 seconds to heat the block to 100C, or 9m46s

This is probably way off but I was bored

[–] SleeplessCityLights@programming.dev 22 points 7 hours ago (1 children)

Your napkin math is the best we have. We will make all decisions based on it.

[–] Napster153@lemmy.world 1 points 3 hours ago

They will have entire hard drives explaining KSP Atlas's shitty math in 3 thousand years...

[–] Contramuffin@lemmy.world 11 points 7 hours ago (2 children)

Hmm, I think at minimum calculus will need to be involved here. Because we can't just assume that the heat is spread evenly in the copper - it'll likely be hotter at the bottom, leading to thermal throttling earlier than expected. On the other hand, there's going to be heat dissipation into the air, which will help cool the block somewhat

[–] Einskjaldi@lemmy.world 2 points 4 hours ago (1 children)

The conduction in copper is fast enough that there's not much of a difference between the top and bottom.

[–] Contramuffin@lemmy.world 2 points 3 hours ago* (last edited 3 hours ago)

Copper conductivity is fast, sure, but it's not fast enough to have equal temperatures at the top and bottom for such a big chunk of copper. That does affect the time to thermal throttle pretty significantly, actually. If we assume completely homogeneous temperatures across the block (ie, instantaneous heat transfer), according to my model, it'll take 703 seconds to thermal throttle. With heat transfer, the time drops to 592 seconds - a difference of about 2 minutes

[–] Eheran@lemmy.world 1 points 3 hours ago (1 children)

Heat transfer will not limit much, but heat loss should add a significant amount of time. How did you model that?

[–] Contramuffin@lemmy.world 1 points 3 hours ago (1 children)

I left another comment going into more detail about the model specifications, if you'd like to read into it. But briefly: I took the copper heat conductivity coefficient and the air heat transfer coefficient. I sliced the copper block into thin slices and modeled heat transfer between each slice, as well as heat transfer between each slice and the surrounding air.

It seems that both heat transfer and heat loss do actually matter quite significantly, but they just cancel each other out almost entirely.

If we assume instantaneous heat transfer, thermal throttling time goes up from 592 seconds to 703 seconds (almost 2 minute difference).

If we assume no heat loss to the air, thermal throttling time goes down from 592 seconds to 500 seconds (about 1.5 minute difference).

[–] Eheran@lemmy.world 1 points 1 hour ago

If they cancel out, the system would be in balance and not get hotter. So some thing does not add up. What heat transfer coefficient did you use and which other numbers etc.?

[–] potpotato@lemmy.world 6 points 7 hours ago (2 children)

Account for convective loses into air?

[–] YerbaYerba@lemmy.zip 4 points 6 hours ago

Its going to radiate a little bit too.

[–] m0darn@lemmy.ca 1 points 5 hours ago

I got ~32 watts of convective heat transfer using assuming a uniform surface temperature of 99°C and a case fan speed of 0.2 m/s

https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html