5 ms·

How can anything be <100% if producing heat is the goal? Where would the rest of the energy be going?

wardedVibe 3y ago
They mean a heat pump. You apply work to transfer heat, getting more inside than if you had just turned the energy into heat directly; the outside gets colder of course. https://en.wikipedia.org/wiki/Heat_pump#Principle_of_operation https://en.wikipedia.org/wiki/Heat_pump#Principle_of_operati...
- josephcsible 3y ago
  Yes, heat pumps are over 100%, but isn't turning the energy into heat directly exactly 100%?
  - Adverblessly 3y ago
    As a random thought, for resistive heating some energy is definitely "lost" as light.
    - krallja 3y ago
      Even light will eventually be absorbed by some surface and become heat.
stephen_g 3y ago
I’m not sure what they mean about crypto miners or computers (mentioning cinebench), basically all the energy going in should be turned into heat so that should be near enough to 100% efficient. Where else would the energy be going? Sure, there will be a bit of RF power (Wi-Fi and some EMI) and depending on the device, a bit of light (LED indicators, monitor backlight) but that should be tiny compared to the energy that will end up as heat… In terms of heating systems though, it’s normal for fuel based systems to be <100% because you need to vent exhaust, and a lot of your heat your fuel produced can be rejected away from where you want it with that. For example, older gas boilers might only be 80% efficient or so, and even the much more efficient condensing boilers are still only in the low 90s.
- crashmat 3y ago
  Won't light heat things up by bouncing around anyway?
  - jffry 3y ago
    Yeah as long as it doesn't escape your room, the photons will end up slightly warming _something_ when they get absorbed.
quesera 3y ago
Using a cryptominer or GP computer at full tilt for the purpose of making heat is less than 100% efficient, because some of the input power gets used on computing. A pure resistive heater is also less than 100% efficient, but it's very very close.
- jffry 3y ago
  Where does the energy "used on computing" go? The power consumption of a CPU is dissipated as heat.
  - quesera 3y ago
    So, I think I understand this view from the thermodynamics sense. First law and all. But by this construction, are all systems which take energy as input (and do not convert it to another form for storage) de facto 100% efficient? When we talk about a (very old) furnace being 75% efficient at turning the chemical energy of a fossil fuel into heat energy, is the 25% loss purely combustion byproducts with some inherent chemical energy plus some non-combusted fuel?
    - jffry 3y ago
      In this case for heating, efficiency is how much of the energy input gets converted into heat in the space you're heating. So for a resistive electric heater, you're dissipating all of the energy as heat (minor quibbles about electromagnetic radiation or status LEDs aside). The same is true for your computer. Burning hydrocarbons have two main sources of energy loss: incomplete combustion, and energy carried away by exhaust gases. Obviously furnace design influences both of these - the US federal minimum is 78%, but high-end furnaces with all the tricks can get over 90% per https://www.energy.gov/energysaver/furnaces-and-boilers https://www.energy.gov/energysaver/furnaces-and-boilers Heat pumps are the third major category, and they are more than 100% efficient because they are using their energy to steal a larger amount of thermal energy from the outdoors air and move it to inside your home (and much of the electric energy they consume is eventually discharged as heat inside your home as well).
  - deleted 3y ago
Rebelgecko 3y ago
I think you lose some to RF, photons (with a monitor), sound waves, etc. Some of those losses can be reabsorbed and turned into heat, but some stray photons and radio waves will go out the window, never to return. The majority does go to heat but you'll never hit the 100% threshold like other methods of heating can meet (or exceed)

They mean a heat pump. You apply work to transfer heat, getting more inside than if you had just turned the energy into heat directly; the outside gets colder of course. https://en.wikipedia.org/wiki/Heat_pump#Principle_of_operation https://en.wikipedia.org/wiki/Heat_pump#Principle_of_operati...

josephcsible 3y ago
Yes, heat pumps are over 100%, but isn't turning the energy into heat directly exactly 100%?
- Adverblessly 3y ago
  As a random thought, for resistive heating some energy is definitely "lost" as light.
  - krallja 3y ago
    Even light will eventually be absorbed by some surface and become heat.

stephen_g 3y ago

I’m not sure what they mean about crypto miners or computers (mentioning cinebench), basically all the energy going in should be turned into heat so that should be near enough to 100% efficient. Where else would the energy be going? Sure, there will be a bit of RF power (Wi-Fi and some EMI) and depending on the device, a bit of light (LED indicators, monitor backlight) but that should be tiny compared to the energy that will end up as heat… In terms of heating systems though, it’s normal for fuel based systems to be <100% because you need to vent exhaust, and a lot of your heat your fuel produced can be rejected away from where you want it with that. For example, older gas boilers might only be 80% efficient or so, and even the much more efficient condensing boilers are still only in the low 90s.

crashmat 3y ago
Won't light heat things up by bouncing around anyway?
- jffry 3y ago
  Yeah as long as it doesn't escape your room, the photons will end up slightly warming _something_ when they get absorbed.

quesera 3y ago

Using a cryptominer or GP computer at full tilt for the purpose of making heat is less than 100% efficient, because some of the input power gets used on computing. A pure resistive heater is also less than 100% efficient, but it's very very close.

jffry 3y ago
Where does the energy "used on computing" go? The power consumption of a CPU is dissipated as heat.
- quesera 3y ago
  So, I think I understand this view from the thermodynamics sense. First law and all. But by this construction, are all systems which take energy as input (and do not convert it to another form for storage) de facto 100% efficient? When we talk about a (very old) furnace being 75% efficient at turning the chemical energy of a fossil fuel into heat energy, is the 25% loss purely combustion byproducts with some inherent chemical energy plus some non-combusted fuel?
  - jffry 3y ago
    In this case for heating, efficiency is how much of the energy input gets converted into heat in the space you're heating. So for a resistive electric heater, you're dissipating all of the energy as heat (minor quibbles about electromagnetic radiation or status LEDs aside). The same is true for your computer. Burning hydrocarbons have two main sources of energy loss: incomplete combustion, and energy carried away by exhaust gases. Obviously furnace design influences both of these - the US federal minimum is 78%, but high-end furnaces with all the tricks can get over 90% per https://www.energy.gov/energysaver/furnaces-and-boilers https://www.energy.gov/energysaver/furnaces-and-boilers Heat pumps are the third major category, and they are more than 100% efficient because they are using their energy to steal a larger amount of thermal energy from the outdoors air and move it to inside your home (and much of the electric energy they consume is eventually discharged as heat inside your home as well).
- deleted 3y ago

Rebelgecko 3y ago

I think you lose some to RF, photons (with a monitor), sound waves, etc. Some of those losses can be reabsorbed and turned into heat, but some stray photons and radio waves will go out the window, never to return. The majority does go to heat but you'll never hit the 100% threshold like other methods of heating can meet (or exceed)