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> Your refrigerator is a heat pump. If you put something warm in there its heat is drawn out to the environment heating your kitchen. I’ve tried to explain to

> Your refrigerator is a heat pump. If you put something warm in there its heat is drawn out to the environment heating your kitchen. I’ve tried to explain to people that putting something warm outside to chill in winter before refrigerating it costs them more in electricity and they look at me like I’m insane.

madacol 4y ago
I don't understand how that conclusion follows from the quote. I still think "putting something warm outside to chill in winter before refrigerating" cost less electricity than refrigerating immediately
- Scoundreller 4y ago
  By outside, I mean outdoors. By bringing "the cold" from outdoors inside, you're going to need an energy source to heat it back up when your home is in heating-mode. Even if you put it inside a small island of cold inside your home. On a net basis, you've made your home colder.
  - madacol 4y ago
    After writing (and deleting) a long response, I finally get your point. You are trying to heat your home, so putting something outside to chill, is to waste heat, but putting it in the fridge would be like recycling that heat from the fridge to your room which should alleviate your room's heater burden.
- bmicraft 4y ago
  If you heat your home using a heat pump, it makes no difference. If you heat with gas, you will indeed use more electricity but less gas when refrigerating immediately.
ianschmitz 4y ago
Because what you said doesn’t make sense
ThunderSizzle 4y ago
Running the refrigerator requires power. Is that power usage more efficient than not running it and relying on another heat source?
- Scoundreller 4y ago
  > Is that power usage more efficient than not running it and relying on another heat source? I can't get whether to answer yes or no to that question, but the fridge's heatpump is more cost-efficient that most people's heat sources. That's the entire premise of the article.
yaantc 4y ago
Parent is right, with some assumptions to be made explicit. Considering the reactions this deserves more explanations ;) Let's say that the heat energy of the warm thing is "E" and the outside and fridge are at same temp for simplicity. Let's also assume that the house is not heated using a heat pump. Consider the two cases: 1) The warm thing is put into a fridge. It's a heat pump, with coefficient of performance C. To cool the thing the fridge will consume E/C, and E will be released as heat inside your room. So you paid E/C of electricity to get E Joules of warming in your home; 2) The warm thing is put outside. The energy E is wasted outside, and then the piece put into the fridge with no extra cooling there. No energy spent by the fridge, but no warming of the house either. To do a fair comparison with the same final state, we need to add E joules of energy to heat the house, which will require E Joule of electricity or primary energy (because no heat pump to heat the house). So to reach the same point, in case (1) we spent E/C of heating energy, and E in the second case. It is indeed more efficient to put the warm thing directly into the fridge, as this will contribute to heating the house with the efficiency of a heat pump. And if you heat your house with a heat pump of same efficient C, you don't care either way: it's equivalent.
- deleted 4y ago
  [deleted]
knolan 4y ago
You’re going to have to explain your reasoning to me there. If your fridge is at 4°C and you put something in there at 21°C it will have to transfer an amount of Joules of thermal energy to the internal environment and then the fridge will have to transport that energy out to the kitchen. The amount of energy depends of the specific heat capacity of the item (J/Kg/K). So to drop the temperature of 1kg of stuff from 21°C to 4°C your heat pump needs to work to remove however many Joules the specific heat capacity determines. If on the other hand if your item is at 0°C then it’s going to heat up via heat transfer from the other items in the fridge. This means the internal temperature of the fridge will drop. But this means the heat pump will have to work less to maintain the internal temperature of 4°C. It’s not that the fridge attempts to heat the fridge up. The fridge simply works less because no fridge is perfectly insulated so it’s easy to let the temperature rise.
- Scoundreller 4y ago
  Because I'm trying to heat my home in winter. Outside = outside_outside = outdoors. Not, uhhh, inside outside like the kitchen counter.

madacol 4y ago

I don't understand how that conclusion follows from the quote. I still think "putting something warm outside to chill in winter before refrigerating" cost less electricity than refrigerating immediately

Scoundreller 4y ago
By outside, I mean outdoors. By bringing "the cold" from outdoors inside, you're going to need an energy source to heat it back up when your home is in heating-mode. Even if you put it inside a small island of cold inside your home. On a net basis, you've made your home colder.
- madacol 4y ago
  After writing (and deleting) a long response, I finally get your point. You are trying to heat your home, so putting something outside to chill, is to waste heat, but putting it in the fridge would be like recycling that heat from the fridge to your room which should alleviate your room's heater burden.
bmicraft 4y ago
If you heat your home using a heat pump, it makes no difference. If you heat with gas, you will indeed use more electricity but less gas when refrigerating immediately.

ianschmitz 4y ago

Because what you said doesn’t make sense

ThunderSizzle 4y ago

Running the refrigerator requires power. Is that power usage more efficient than not running it and relying on another heat source?

Scoundreller 4y ago
> Is that power usage more efficient than not running it and relying on another heat source? I can't get whether to answer yes or no to that question, but the fridge's heatpump is more cost-efficient that most people's heat sources. That's the entire premise of the article.

yaantc 4y ago

Parent is right, with some assumptions to be made explicit. Considering the reactions this deserves more explanations ;) Let's say that the heat energy of the warm thing is "E" and the outside and fridge are at same temp for simplicity. Let's also assume that the house is not heated using a heat pump. Consider the two cases: 1) The warm thing is put into a fridge. It's a heat pump, with coefficient of performance C. To cool the thing the fridge will consume E/C, and E will be released as heat inside your room. So you paid E/C of electricity to get E Joules of warming in your home; 2) The warm thing is put outside. The energy E is wasted outside, and then the piece put into the fridge with no extra cooling there. No energy spent by the fridge, but no warming of the house either. To do a fair comparison with the same final state, we need to add E joules of energy to heat the house, which will require E Joule of electricity or primary energy (because no heat pump to heat the house). So to reach the same point, in case (1) we spent E/C of heating energy, and E in the second case. It is indeed more efficient to put the warm thing directly into the fridge, as this will contribute to heating the house with the efficiency of a heat pump. And if you heat your house with a heat pump of same efficient C, you don't care either way: it's equivalent.

deleted 4y ago
[deleted]

knolan 4y ago

You’re going to have to explain your reasoning to me there. If your fridge is at 4°C and you put something in there at 21°C it will have to transfer an amount of Joules of thermal energy to the internal environment and then the fridge will have to transport that energy out to the kitchen. The amount of energy depends of the specific heat capacity of the item (J/Kg/K). So to drop the temperature of 1kg of stuff from 21°C to 4°C your heat pump needs to work to remove however many Joules the specific heat capacity determines. If on the other hand if your item is at 0°C then it’s going to heat up via heat transfer from the other items in the fridge. This means the internal temperature of the fridge will drop. But this means the heat pump will have to work less to maintain the internal temperature of 4°C. It’s not that the fridge attempts to heat the fridge up. The fridge simply works less because no fridge is perfectly insulated so it’s easy to let the temperature rise.

Scoundreller 4y ago
Because I'm trying to heat my home in winter. Outside = outside_outside = outdoors. Not, uhhh, inside outside like the kitchen counter.