Is heating frozen peas more energy-efficient with or without a lid on the pot?

Question

Is it more energy-efficient to heat frozen peas on the stove in a pot with or without a lid ?

Although I'm a complete physics dummy, I am trying to contribute some food for thought. Sorry if the following arguments are flawed or even stupid:

potential arguments contra lid:

1. The lid prevents the cold from escaping the pot.
2. It also prevents outside air at room temperature from entering the pot and helping with thawing.

potential argument pro lid:

1. The lid prevents the heat generated by the stove from escaping the pot.

potential argument for a lid/lidless-combination:

1. Given that I pour peas at -18°C into the pot, I assume the initial temperature in the pot is colder than outside. Maybe that's why a lid/lidless-combination would work best: lidless at the beginning as long as the inside temperature is colder than outside, and once the former exceeds the latter, I need to put the lid onto the pot ?

assumptions and test setup:

• standard room temperature of 20°C
• standard freezer temperature of -18°C
• In case the type of stove is decisive in answering the question (which I doubt), let's assume an induction stove.
• The target temperature should be "warm" or "appropriate for serving a warm meal in a restaurant". (If this is too unspecific, let's assume a safe target temperature of 60°C.)
• I do not use any added water for cooking. (I merely put a tiny amount of oil into the pot, so that the peas don't stick to the pot. I could as well use a pot with a non-adhesive surface instead of the oil.) However, the process of thawing sets free a bit of water contained in the peas. Thus, heating the peas does produce some steam, despite no added water being used.

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related questions:

• Energy in bolognese reduction - lid on or off?
• Time taken to cook frozen peas
• Does placing a lid on a pot affect the rate of boiling?

Answer 1

The lid prevents the cold from escaping the pot.

"Cold" does not go anywhere. If you touch an ice cube for example, you are not feeling "cold travel" to your finger but heat moving from your finger into the ice.

With the lid off, the higher temperature of the air will allow for some heating of the peas, but not anywhere near as much as the heat generated by the stove.

It [lid on] also prevents outside air at room temperature from entering the pot and helping with thawing.

The temperature inside will be greater than that outside, so it is the heat from the source that will (through conduction and convection) play the greatest role in the thawing process.

The lid prevents the heat generated by the stove from escaping the pot.

Yes.

With the lid on, thermal energy of the water and steam remain in the pot with little leaking to the environment. This will allow the peas to heat up faster and the stove need not do as much work as it would with the lid removed. So it's quicker (using less energy) as well as loosing less energy to the environment.

So, it is more efficient to leave the lid on.

Answer 2

Let's model the heat transfer as follows:

1. Heat flows from the stove element into the metal of the pot via conduction.
2. Heat flows from the metal of the pot into the water/ice mixture via conduction. (We'll not distinguish between the peas versus the ice crystals they contain.)
3. Heat is transported within the water/ice mixture via convection.
4. Heat is exchanged between the water/ice mixture and the air via conduction at the surface.
5. Heat circulates within the air in the top half of the pot by convection. (Once the water boils, there will also be mass transport into the air in the top half of the pot, and the hot steam will carry heat with it as well.)
6. Heat can be exchanged with the air in room by convection if there is lid. But if there's a lid, the heat can only escape to the room by conduction through the lid.

Heat energy always flows spontaneously from hot places to cold places. The most efficient way to heat the peas is to arrange each of steps so that heat is flowing towards the peas, rather than away.

When you first put the frozen peas in the water, the temperatures of the peas and of the water will both more towards zero degrees. (If there's way too much very cold ice, you could in principle freeze all the water; what's more likely is that the ice component reaches zero degrees first and the room-temperature water would eventually melt the ice.) Without the stove on, this cold water/ice mixture would produce a layer of cool air above it. Heat would be transferred more efficiently from the room to the water/ice mixture if there were no lid, so that convection (including random fluctuations) could replace this cold air above the water/ice mixture with warmer air from elsewhere in the room.

With the stove on, the water will eventually get hotter than the room temperature, and so will the layer of air in the top half of the pot. At this point, you want to keep that warmer air nearby, to reduce the heat loss from the water to the room. That's when you put the lid on.

So the efficient method is:

• put the peas in the water in the pan on the stove
• turn the stove on
• put your hand just above the water's surface and notice that the air is cold
• twiddle your thumbs for a minute
• put your hand just above the water's surface and notice that the air is warm
• put on the lid

The efficiency gain is really tiny, so you might also just not worry about it.

Answer 3

If we consider the question to be more broadly “Is it better to allow all heat transfer modes while they are, separately and in conjunction, still transferring the most net energy to the target?”, then I agree with your intuition.

@josephh makes correct points that it’s more useful to think in terms of heat transfer than “cold transfer,” and that the heat source in a stove is strong, and that bottom heating soon produces hot air and steam that heat the target more than the surrounding air (and would wastefully escape without the lid).

We can reconcile all of this by supposing that it provides a slight heating boost if the lid is kept off for just the first few seconds, and if you fanned room-temperature air around the colder target, say, until the bottom of the vessel is hot. In other words, the difference is minimal in practice for the example you’ve chosen. But I don’t want to discourage your intuition regarding the broader point.