If I waste 6 ice-cubes (typical US refrigerator) - how many hours is that equivalent of leaving a 13W light bulb on ? Consider an EPA certified high efficiency refrigerator and the ambient temperature to be 75F
Thanks
If your ice cubes are 20g, then 6 are 120g. 120g of water, starting at a room temperature (25C), requires (25 + 80) calories/g to freeze, and then another 5/g to get it to -5C. So, 13,200 calories all told. If the refrigerator is 30% efficient, then it will consume ~40,000 calories to freeze your cubes.
1 calorie is equal to 1.16222222 × 10-6 KWH, so freezing your cubes will consume .0464 KWH. A 13W lamp consumes .013 KWH/H, so you could run your 13W lamp for 3.57 hours on the energy you wasted by throwing the ice cubes away.
(I had to guess on the refrigerator efficiency).
And unfortunately you guessed poorly. The coefficient of performance of a refrigerator is typically going to be more than an order of magnitude larger than that. Yes, it’s above one: This is possible because it’s just moving energy around, not producing it itself.
It’s hard to calculate. During the heating season the heat cost is zero, with all the heat transfer warming your kitchen, and the efficiency losses doing the same. During the cooling season the same phenomena applies, but you’ve got the added expenditures of having to use your air conditioning to move the heat once more to the outside.
Only for heating, not for cooling.
No, it works for any heat pump. The measurement in question is the coefficient of performance and could be 5 or more for a typical freezer (roughly corresponding to 500% “efficiency”).
That said, I’m finding it hard to find actual real-world CoP numbers for typical refrigerators. This paper implies that they’re around 1.0 in practice. Maybe someone else has access to better figures.
OK then.
For a freezer with COP = 1, my numbers get divided by 3, so the lamp could burn for 1.2 hours or so.