heat energy – Thoughts on Energy

Melting Snow while Backpacking

Introduction

I enjoy backpacking in the great outdoors. As all good backpackers know, planning for access to water is a critical part of any backpacking trip. Backpacking is a lot of work and usually involves a lot of sweating, so keeping hydrated is necessary. However, maintaining a backpack as light as possible is also a key consideration, so one doesn’t want to carry a lot of water around. Thus, the usual strategy is to stay close to natural sources of water and bring a good water filter to allow for regular refills of water containers.

As a side note, how much work is a backpacking trip, you ask? Oh, at 100% efficiency, almost one million Joules worth of work would be required on a typical mountainous trail with about 1,000 meters of up and down (average 5% grade for a 20 kilometer, or about 12.5 mile, trip) during the full trip. (That calculation is simply the potential energy, which is the mass multiplied by acceleration due to gravity multiplied by the total elevation change). However, considering the body is probably only about 5% efficient in this case, the real energy use is approximately 20 million Joules, which is equivalent to a total of 5,000 food calories. That is just the extra calories of food energy “burned” during the trip in addition to the resting metabolic calorie burn of about 2,000 calories per day.

While on a backpacking trip some time ago, my brother and I took a day hike up a mountain and we knew we would not be near any sources of water for a good portion of the day. I had two water bottles with me on the backpacking trip, one un-insulated bottle and one insulated Thermos bottle. However, I didn’t have enough foresight to bring a smaller bag on the trip to carry around for the day hike, so all I had was a plastic grocery sack (I know, some Californians reading this post will cringe at those words), and thus I only took one of my water bottles. Unfortunately, it turned out that I selected the wrong water bottle to take with me.

You see, even though I tried to ration my water intake, I found I was out of water with several miles left to hike. Fortunately, we found some snow that I ate a good amount of, and also packed a good amount into my water bottle. I figured it could melt as we went along our merry way and I would soon have some nice cool water to drink. Well, that didn’t work out as well as I hoped because the snow was painfully slow to melt. Ironically, as we continued our hike, we ended up in a downpour that lasted a few hours. So I was soaked to the bone and dehydrated at the same time, and I still couldn’t get the darn snow to melt in my water bottle!

I had plenty of time during the hike to think about methods to melt the snow into water so I could drink it. Since it relates to energy, I thought I would share my findings here. The underlying assumption below is that conduction heat transfer through the insulated water bottle was negligibly low in the short term, which I think turned out to be a pretty good assumption. Also, a spreadsheet containing the calculations is available below.

Air Blowing Method

My first thought was to simply blow air into the water bottle’s straw and have it come out the little pressure release hole. It would work like any basic heat exchanger. I could transfer heat from the air I exhaled into the snow and cause it to melt. We should all be somewhat familiar with heat exchangers, as many of us have at least two in/near our dwelling places as part of an air conditioning system. One heat exchanger (located inside) allows the transfer of heat from the air inside the living space into a refrigerant, and the other heat exchanger (located outside) allows the transfer of heat from the refrigerant to the outside air.

We can use a simple equation to determine how much air I would need to blow through my water bottle straw to melt the snow. To formulate this equation, I simply use the fact that the amount of heat required to melt the snow must be the same as the amount of heat lost by the air as it travels through the snow in the bottle.

Where

Q is the total amount of heat required to be transferred from the air to the snow, in Joules.
is the total mass of air required, in kilograms.
is the specific heat of air, which is 1.0 Joules per kilogram degrees Celsius.
is the change in temperature experienced by the air as it moves from my mouth to the exit hole, in degrees Celsius. I assume the air temperature goes from about 32 degrees Celsius when it exits my mouth to about 5 degrees Celsius when it exits the water bottle, for a change in temperature of 27 degrees C.
is the total mass of snow, in kilograms. My water bottle fits about 0.5 liters of water. I haven’t packed the snow in completely tight, so I have about 0.4 liters of snow at a density of about 0.6 kilograms per liter, for a total of 0.24 kilograms of snow.
is the latent heat of fusion ((Wikipedia link)) of water, which is 334,000 Joules per kilogram.

Now all we need to do is solve for to obtain how much air it would take going through the “heat exchanger” to melt the snow. However, just knowing the total mass of air required is not very helpful. What I really want to know is how long it will take to get that mass of air through the water bottle. Thus, I need to know my respiration rate. I need to know how much air (in units of mass, or kg) I can blow through the bottle in a given amount of time. A typical adult respiration rate is about 6 liters of air per minute. However, when I am hiking with a backpack on I am exercising relatively vigorously, so I probably have a respiration rate around 50 liters per minute.

At an altitude of about 6,000 feet above sea level, the density of air is about 1.0 kilograms per liter. Using all of this information and solving for the time required to melt the snow (see the spreadsheet attached below if you want to follow all the math there), I obtain:

So it would take me about an hour to get enough air through the water bottle to melt the snow. However, this assumes that all of the air I exhale goes through the water bottle. I found that to be unsustainable in practice, as the water bottle restricted my exhalations enough that I would quickly run out of breath (remember, I am hiking on mountains wearing a 35-pound backpack). So I could really only get about one third of the air I exhaled through the water bottle. That results in about three hours to melt the snow. I am not that patient.

Potential Energy Input Method

My next thought was to use potential energy of the snow to melt it. The thought process is relatively simple. When I held the water bottle upright, there was a small pocket of air at the top. So if I turned the water bottle over, there would be stored potential energy in the lump of snow because the snow tended to stick to the bottom of the bottle.

As you will recall, potential energy is stored mechanical energy. In this case, it is the energy of the snow that can be realized as the snow falls the short distance from one end of the water bottle to the other. That potential energy must go somewhere as the snow falls. Initially, some energy becomes kinetic energy as the snow moves. When the snow stops against the bottom end of the bottle, about the only place for that energy to go is in the form of heat to the snow. Bingo! All I have to do is tip the water bottle upside-down enough times and I convert enough potential energy into heat energy to melt the snow!

So how many times do I need to invert the water bottle? Using the equation from the previous example, I already know it takes about 80,000 Joules of energy to melt the snow. So I just need to figure out the amount of potential energy for the snow in one time of inverting the water bottle.

It turns out the potential energy is about 0.1 Joules. So I only need to invert the water bottle 850,000 times. Assuming I can invert the water bottle once per second, this ends up being about 10 days. I am definitely not that patient.

Shaking Method

So what if I could speed up the potential energy process with some kinetic energy instead? Instead of inverting the bottle and waiting for the snow to fall in the water bottle, I can just orient the bottle horizontally and shake it back and forth instead and get a similar effect. I can increase the rate of impacts between the snow and the inside of the water bottle and maybe even increase the amount of heat added to the snow on each impact.

Consider the physics of what is going on inside the bottle as I shake it back and forth. As I shake the bottle in one direction, the bottle and snow move together. Then as I move the bottle in the other direction, the snow slams into one end of the bottle. This motion can be represented as a sinusoidal function. All I need to know are the amplitude of the motion and the period of the motion and I can fully describe the maximum velocity and acceleration of the motion. I assume the total range of motion is about 20 centimeters and I can move the bottle back and forth about three times per second. Thus, the motion of the bottle can be represented in this plot:

The position, velocity, and acceleration of the bottle can be represented by these equations (with just some calculus magic going on in the background):

In these equations, A is the amplitude of the motion, which is 10 centimeters, and T is the period of the motion, or the time it takes to complete a full cycle of motion, which is 0.333 seconds. The maximum velocity of the bottle (represented by the first portion of the velocity equation) is about 2 meters per second and the maximum acceleration (again, represented by the first portion of the acceleration equation) is about 35 meters per second squared, or about 3.5 times the acceleration of gravity. Considering a major league baseball pitcher can move a baseball from zero to 95 miles per hour in about 0.3 seconds (acceleration of about 15 times that of gravity), that seems like a reasonable value for acceleration.

It is the maximum velocity that I am interested in since that feeds into the calculation of kinetic energy for each cycle of bottle motion. It turns out that the kinetic energy in one impact of snow using the shaking method is about four and a half times the potential energy of the snow using the gravity method. In addition, the rate of impacts is about six times the rate for the gravity method. Thus, the shaking method is about 27 times as efficient as the gravity method. Still, it would take about nine hours to melt the snow using the shaking method, and that is assuming my arms don’t fall off before that point as a result of doing that much shaking. Nine hours is still too long.

Conclusions

In conclusion, none of the methods I tried turned out to be very effective for melting snow inside the insulated water bottle. The amount of time calculated for each method is shown in the table below. For comparison, I considered how long it would take the snow to melt if I had an uninsulated container sitting in room temperature (with heat transfer via convection of the surrounding air), using similar analysis to a previous post. It turned out that that method would actually result in a shorter time to melt the snow than any of the other methods.

So how did I end up melting the snow? The truth is, I never got the snow fully melted during the backpacking trip. When I found that the snow was not melting in spite of my best efforts, I decided to try another experiment. I decided to see how long it would take the snow to melt inside the bottle. Sure enough, when I arrived home approximately 24 hours after I put the snow in the bottle, there was still a small amount of snow remaining. So an insulated water bottle is pretty effective at keeping a drink cold. And if you are out backpacking and see some snow you want to use for drinking water, I recommend using a fire to melt it!

If you are interested in the calculations, see this spreadsheet:

Backpacking Snow Melting Calculations Download

Should I Leave the Milk Out?

Introduction

For someone like me who constantly thinks about energy, even getting cold cereal ready to eat in the morning can be a mind-engaging task. As I pull the milk out of the refrigerator to pour on my cold cereal, I know I will probably want the milk again after I finish my cereal in order to pour myself a bowl of fresh milk to finish off my breakfast. The question always pops into my head, “should I leave the milk out, or put it back in the fridge until I need it again?” Of course, the correct answer to this question is, it doesn’t really matter. But for an energy enthusiast like me, I want the real answer. The answer that involves the least amount of energy wasted.

To reiterate, the two options, with respect to what happens with the fridge and the milk, are:

Option A: Open fridge (assume the fridge is open for about 5 seconds each time it is opened) and take milk out, leave the milk on the counter for 15 minutes, and then open fridge to return the milk.
Option B: Open fridge, close fridge for about 15 seconds while pouring milk, open fridge again to return the milk; 15 minutes later, open the fridge again to get the milk out, close fridge for 15 seconds while pouring milk, open fridge to return the milk.

So, the way I see it, it comes down to which option wastes more energy. Both options involve some “waste” of energy from the fridge when the fridge is opened to remove and return the milk. The “extra” energy wasted in Option A is the loss of coldness of the milk sitting on the counter for 15 minutes, while the extra energy wasted in Option B is the extra loss of cold air from the fridge when it is opened two additional times. So we just need to calculate the wasted energy.

We will make some assumptions as we go through these calculations, and the answers will vary based on the assumptions we make and the uncertainty in the numbers we use. Keep in mind that for this particular problem, we aren’t really looking for exact answers, we are looking for good enough answers that will allow us to compare the two options. Do both options result in about the same energy waste, or is one of them substantially greater than the other?

Another thing to keep in mind is that we aren’t just looking for the answer to this specific question, we are thinking about how this answer will apply to similar situations. What if what we are removing from the fridge isn’t milk, but something else? What if we are removing something from the freezer instead of the fridge? What if that something is out for an hour instead of just 15 minutes?

While we perform our calculations and compare values, we are looking at the broader picture, for rules to govern our use of the refrigerator in general. Can we come up with some good general ground rules and energy estimates that will make our future use of the refrigerator as efficient as possible? That is the goal.

Picture of a refrigerator packed with various contents, including milk. — Opening the fridge is always an adventure…

Internet Research

Now, before we go performing a bunch of calculations, we will check to see what the Internet says about this energy efficiency question. The topic that usually comes up under any searches about whether I should leave the milk out is food safety. The standard answer for how long milk can be left out of a refrigerator without spoiling seems to be about two hours. However, some Europeans normally leave their milk out of the fridge due to a different pasteurization process, though the percentage of milk of this type of milk that is consumed varies from country to country.

But that isn’t what we are looking for about energy efficiency. It turns out there are some ideas out there about how long the refrigerator door should be left open, such as this insightful question and answer post. There are also plenty of general energy saving tips for refrigerators and other kitchen appliances, but they don’t generally get into the details of how long or how often doors should be left open.

Calculation for Option “A” Energy Wasted

So it looks like we will just have to calculate the answers and see how things turn out. In the calculations below, I am assuming a full or nearly full gallon of milk. Let’s start with the extra energy wasted in Option A: the energy wasted with the milk sitting out on the counter. When the milk container sits out for 15 minutes, there are four ways that heat is being transferred into the milk:

Convection heat transfer from the surrounding air,
Conduction heat transfer through the surface the container is sitting on,
Condensation of water vapor on the outside surface of the container, and
Radiation heat transfer from the surrounding environment.

A graphic depicting four mechanisms by which heat is transferred into the milk and container: convection, conduction, radiation, and condensation. — Four mechanisms by which heat is transferred into the milk and container.

It is instructive to first consider the total amount of heat transfer required to occur to bring the entire gallon of milk up to room temperature. To calculate this, we simply use the heat equation:

Heat equation: Q equals m c delta T. Heat energy equals mass multiplied by specific heat capacity multiplied by the temperature difference.

The specific heat capacity of milk is approximately 3,790 joules per kilogram degrees Celsius, and one gallon of milk has a mass of 3.8 kilograms. Thus, the heat required is:

We can expect the heat transferred in 15 minutes to come well short of this value, as we know by experience that it takes at least a few hours for milk to get even close to room temperature. In starting my heat transfer calculations, I can immediately discount the effect of radiation heat transfer, as I know it is relatively very small in cases with relatively minimal difference in temperature (which would include this case with only 20 degrees Celsius difference).

Thus, I start by calculating convection heat transfer. Convection heat transfer calculations get fairly involved, so I will spare you the gory details of a slew of dimensionless numbers (including the Prandtl number, the Grashof number, the Nusselt number, the Rayleigh number, the Biot number, and the Fourier number) and strange correlations (including raising a factor to the power of 8/27). If you are interested, though, you can see my spreadsheet with the calculations attached below.

Suffice it to say, a good estimate for heat added to the milk via convection during the 15 minutes sitting on the counter is about 15,000 joules.

I can now calculate the conduction through the surface. I do this by using equations that assume the surface that the milk is sitting on is a semi-infinite solid. This assumption is reasonable in some situations but not others. I could attempt to use finite element analysis and numerical methods to come up with better answers, but instead I am just going to use engineering judgment to modify the answers into something reasonable based on what I know about the scenario. Anyway, it turns out that it matters a great deal what material the surface is made of. The table below shows the calculated energy transfer due to heat conduction in 15 minutes for different surfaces, along with my values modified to be more realistic.

Table of energy values in joules for different counter surfaces: steel at approximately 22,000 joules, granite at 12,500 joules, wood at 3,000 joules, and hot pad at 900 joules.

As the calculation indicates, the heat loss when the milk is sitting on steel is much greater (over 150 times greater!) than the heat loss when the milk is sitting on a hot pad, towel, or equivalent insulator. My modifications to make the values more realistic are based on the fact that the materials are not actually semi-infinite (there is a limited volume of material for the heat to be transferred into, which is particularly significant for the steel surface) and to account for the thermal resistance of the milk container and its contact with the surface. At any rate, a reasonable average value for heat loss considering a wide variety of counter top surfaces is approximately 5,000 joules.

Next, I can calculate the heat added to the milk as a result of condensation. The energy transferred as a result of condensation is simply the latent heat of vaporization of the mass of water being condensed from water vapor into liquid water on the surface of the milk container. Of course, the amount of condensation will vary significantly depending on the humidity content of the air. I made some reasonable estimates of the amount of condensation that will occur in my calculations (see full details in spreadsheet attached below) and came up with a figure of approximately 5,000 joules of heat transfer due to condensation.

So, in summary, for Option A, the wasted energy is:

Table showing wasted energy via four heat transfer mechanisms: convection at 15,000 joules, conduction at 5,000 joules, condensation at 5,000 joules, radiation being negligible for a total of 25,000 joules.

In case you are interested, see attached file for details of heat transfer calculations for this section:

Milk Calculations Download

Calculation for Option “B” Energy Wasted

Next, we can calculate how much energy is “wasted” when the fridge is opened. The “waste” that occurs when a refrigerator is opened is a result of the cold air inside the fridge being replaced with room temperature air from outside the fridge. The total interior volume of a typical fridge is approximately 0.3 to 0.5 cubic meters. However, some of that volume is taken up by the contents of the fridge, so maybe a good average number for air volume of a fridge is 0.25 cubic meters. The mass of that air is simply:

Mass of air equals volume multiplied by density.

V is the volume of the air in the fridge and ρ is the density of the air, with a typical value being 1.2 kilograms per cubic meter. Thus, the mass of air in a fridge is about 0.3 kg. To calculate the energy required to cool this air from room temperature (about 22 degrees Celsius) to fridge temperature (about 2 degrees Celsius), as in the previous section, we simply use the heat equation:

The specific heat capacity of air is approximately 1000 joules per kilogram degrees Celsius. Thus, the heat required is:

Another factor to consider is the water vapor within the air that enters the fridge. While humidity levels in the fridge and in the air outside the fridge can vary significantly, let us examine a typical scenario just to get an idea.

Say the air in your home is at 50% relative humidity and the air in the fridge is at 100% relative humidity (not an uncommon situation if the fridge has been opened recently or if there is anything wet in the fridge). According to this chart, that would mean replacing air in the fridge with 0.005 pounds of water per pound of air with air from the room with 0.008 pounds of water per pound of air. This is an extra 0.003 pounds of water per pound of air. Since we have 0.3 kg of air, that means approximately one gram of extra water vapor in the fridge.

That small amount of water vapor doesn’t seem like a big deal. However, it takes a relatively large amount of energy to condense that one gram of water vapor, about 2,000 joules in fact. If the air in the house is hot and/or humid, the energy required to condense the additional water vapor can exceed the energy required to cool the air itself.

So all together, we can estimate that the energy “wasted” when the air in the fridge is replaced with room temperature air is approximately 10,000 joules. Of course, that represents the situation in which the entire volume of air from the fridge is totally replaced with room-temperature air. Since the fridge door is only open for 5-10 seconds, maybe only one quarter to one third of the air is replaced. So, we can estimate that one fridge door opening is approximately 3,000 joules of wasted energy. So two extra fridge door openings result in approximately 6,000 joules of wasted energy.

(Note that in our scenario, both extra door opening occur just 15 seconds or so after the previous opening, so much of the energy loss due to opening the door would have already occurred. That is, the air in the fridge is already somewhere between normal fridge temperature and room temperature. Thus, the amount of energy waste is actually some value lower than 6,000 joules for the two extra fridge openings, but the 6,000 joules is still a reasonable estimate.)

Conclusions

For the specific scenario postulated in the introduction, leaving the milk out of the fridge for 15 minutes wastes more energy than opening the fridge door two extra times to put the milk back and get it out again (25,000 joules compared to 6,000 joules).

If the period of time for the milk to be out was reduced to four minutes, it would be a close call as to which option would waste more energy (both options wasting about 6,000 joules in that case). So four minutes is probably a good cut-off time. If a full gallon of milk will be out for any longer than four minutes, it is better to just open the fridge and put it back in, then get it out again later.

I should note here that the energy waste we have discussed above is not the same as the extra electrical energy required to run the refrigerator. Refrigerators have a typical coefficient of performance of three to four, meaning that for every joule of electrical energy used by the refrigerator, three to four joules of heat energy can be removed from the interior of the fridge. So for scenario above with 6,000 joules of wasted energy, the fridge may use about 2,000 joules of electrical energy. That is about the energy required to power a 60-watt light bulb for about 30 seconds. (Yes, I realize most “60-watt” light bulbs do not actually take 60 watts of electricity to run anymore due to the efficiency of compact fluorescent and LED bulbs).

In the end, perhaps opening the door of the fridge isn’t as bad as we might think in terms of energy efficiency. Once the fridge has been open for a while, though, the cooling system will kick on and the cool air will mostly be wasted immediately to outside the fridge. Since a typical fridge cooling system uses 100-125 watts to run, it only takes about 30 seconds of the fridge running to remove the amount of heat equivalent to replacing all the room temperature air in the fridge with normal fridge temperature air. However, if the fridge is left open for a while, all of that 100-125 watts of the fridge running is almost immediately wasted as cold air flows right out the door.

I should also note that the amount of energy wasted by a refrigerator varies with the season of the year. In the winter time, when energy is being used to heat a home anyway, energy used by the fridge to cool the interior space is put out into the home as heat. So the energy isn’t really wasted at all. A person could, in fact, heat their home just by leaving the door of the refrigerator open all the time, but this isn’t really recommended (as the food wouldn’t stay cold and the fridge probably wouldn’t last long running all the time). Also, for those that have natural gas furnaces, using the furnace to heat the home is much more efficient than using electricity (such as using a refrigerator).

However, in the summer time, when air conditioning is on, the energy wasted by the fridge causes even more energy waste than just that used by the fridge itself. The air conditioning system must work to remove the heat energy (which has been converted from electrical energy by the fridge) from the home. So if you are going to pick a time of year when you are really strict about saving energy for the refrigerator, pick summer time.

Perhaps the most notable energy efficiency tips for use of a fridge that are not immediately obvious are the following:

There is value in keeping the air in the fridge dry by not leaving standing water or wet, uncovered foods because it takes a relatively large amount of energy to condense water (compared to the relatively small amount of energy to cool air).
When items are removed from the fridge, consider placing them on a hot pad or towel instead of directly on a counter or table surface (especially if that surface is metal!).

In conclusion, I hope this post has spurred some thoughts about energy efficiency for refrigerators. I hope I have answered the question “should I leave the milk out?” Of course, while each of us acting individually can’t save a whole lot of energy based on how we use our refrigerators, maybe if we all tried to be just a little more efficient, our combined efforts would add up to a big impact in terms of less waste and a better overall world environment. That is the dream, anyway.

And one final note. If you enjoyed this post, you might also enjoy a somewhat related post that answers the question of what appliance wins between a freezer and a toaster. Enjoy.

Measuring Energy

How is energy measured? This is a great question, and deserves our attention in this post. As we consider how energy is measured, we will run into the most commonly used units of measurement for energy, joules (J), British thermal units (Btu), and watt-hours (Wh).

It turns out that measuring energy is difficult. Energy is not something that is readily accessible to our five senses. We can’t see, hear, touch, smell, or taste energy, at least not directly. Some properties of the world around us are readily available to our five senses, such as distance, mass and temperature. We can see physical dimensions that something has and can easily compare the size of one object to another. We can feel whether something is hot or cold.

Though we may not be able to see or touch something and immediately convert it to an exact dimension (as in, “I can see that this smartphone is exactly 15.3 centimeters long” or “I just walked outside and I can feel that the air temperature out here is exactly 10.7 degrees Celsius”), we can usually get pretty close (“this smartphone is between 10 and 20 centimeters long” or “the outside temperature is between 5 and 15 degrees Celsius”), and can almost always compare things like distance, mass, and temperature to know what things are bigger or hotter.

The same is not true of energy. We do not have an intuitive sense for energy like we do for other properties. Energy takes higher-level thinking. Consider these three examples:

What takes more energy, lifting a 20-pound bag of potatoes from the floor to the kitchen counter or heating a small pot of water to boiling temperature?
What takes more energy, driving your car to the grocery store or vacuuming the floors in your entire house?
What releases more usable energy: one pellet of nuclear fuel (the size of an average marble) in a reactor in a nuclear power plant or 1,000 pounds of coal in a coal power plant?

Did you come up with the answers right away? How confident are you in your answers? Maybe you didn’t even hazard a guess because you just have no idea. I will share the answer to the first question later in this post (and the other two in a future post), but for now, let us consider why these questions may seem so difficult.

As mentioned, energy is not intuitively accessible to our immediate observations of the world around us. Thus, it wasn’t until the 1800s that the word “energy” was actually used in its modern sense of meaning, as a physical characteristic of an object or system that can actually be calculated and quantified. While it would be fascinating to dig into the history of the concept of energy, that is not my purpose here.

In order to understand how energy is measured, calculated, and quantified, it will be instructive to look at the meaning of some of the most widely used units of energy.

Units of Energy

The most widely-used unit of energy worldwide is the joule, named after James Prescott Joule, an English amateur scientist who lived from 1818 to 1889. A joule is equal to one kilogram meter squared per second squared, or one newton-meter. It is the amount of energy involved in moving with a force of one newton over a distance of one meter. The joule serves brilliantly within the overall International System of Units (also known as SI units) to measure various types of energy, including mechanical energy (kinetic or potential), chemical energy, and heat energy. The use of the joule is often the simplest method for completing energy conversions and calculations.

A watt-hour is a unit of energy that is often used to describe energy that is used over some period of time. A watt is a unit of power (amount of energy used over a given time period), defined as utilizing one joule of energy every second. A watt-hour is the amount of energy required to provide one watt of power over one hour of time. Since there are 3,600 seconds in an hour, one watt-hour equals 3,600 joules.

A British thermal unit (Btu or BTU) is the amount of energy that is needed (when added as heat) to raise the temperature of one pound of water by one degree Fahrenheit. The Btu is typically used (instead of the joule) in engineering calculations involving heating or cooling in the United States.

A separate unit in the English system of units is the foot-pound. It is defined in a similar manner as the joule, with one foot-pound being equal to the energy involved in moving with a force of one pound-force over a distance of one foot. However, the use of a pound-force can create confusion because the term “pound” is utilized to describe both mass and force in the English system of units. The foot-pound is used when mechanical energy (kinetic or potential energy) or work is involved.

Let us now use the above units to solve the first energy comparison question mentioned above: what takes more energy, lifting a 20-pound bag of potatoes from the floor to the kitchen counter or heating a small pot of water to boiling temperature?

Energy Comparison: Energy in the Kitchen

In the calculations of energy below, it is important to consider what quantity of energy is actually being calculated. In almost every process where there is a conversion of energy from one form to another, there is some inefficiency in the process. When lifting a bag of potatoes, some type of energy (if a person is doing the lifting, chemical energy in the form of glucose in the muscles or if a machine is doing the lifting, stored or produced electrical energy) is converted into potential energy as the bag is lifted up against the force of gravity.

The conversion of energy is never perfectly efficient. So when we talk about energy, we need to be careful and deliberate about what quantity we are actually measuring or calculating. Are we calculating the energy to perform a task in an ideal (imaginary) system or in the real world? The examples below will help to illustrate the difference.

So, what takes more energy, lifting a 20-pound bag of potatoes from the floor to the kitchen counter or heating a small pot of water to boiling temperature? First, we will determine the energy required to lift a 20-pound bag of potatoes to the kitchen counter.

As noted above, there is some inefficiency in the human body converting chemical energy stored in food to mechanical energy (work of physically lifting the bag of potatoes). In performing this energy calculation, we are ignoring that inefficiency and just considering the ideal (minimum) amount of energy to lift the bag of potatoes. That is, if the system for lifting the potatoes was perfectly efficient at converting stored energy into work to lift the bag of potatoes, how much energy would that system use?

Considered in this way, the energy required to lift the bag of potatoes is simply the difference in potential energy of the bag of potatoes between the countertop and the floor. The equation is: the difference in potential energy equals the mass multiplied by the acceleration of gravity multiplied by the change in height:

Let us first solve this equation using English units:

The result is 2,600 pounds-mass square feet per second squared. However, this unit is somewhat unwieldy. The more customary way to represent this amount of energy in the English system of units is using the unit of foot-pound. To understand how force and mass and mechanical energy are expressed in the English system of units, one would need to delve into the details of the units of the slug, pound-mass, and pound-force. Needless to say, it gets confusing. At any rate, expressed in foot-pounds, the energy would be 80 foot-pounds.

Now let us solve this equation using SI units:

A good approximation can be determined in this case without even using a calculator. Twenty pounds is approximately ten kilograms, and four feet is approximately a meter. The acceleration of gravity in SI units is about ten meters per second squared. So it is easy to calculate the energy in your head in this case and come up with about 100 joules, which would be pretty close to the right answer.

Now, we will determine the energy required to heat a small pot of water to boiling temperature. First, we need to make some assumptions. We will assume a “small pot” is about one liter (or 1.06 quarts) of water, and the temperature of the water is 50 degrees Fahrenheit (or 10 degrees Celsius).

Similarly in this calculation, there will be some loss of efficiency. If the heat source is a burner using a fuel (natural gas or propane), the gas will not burn perfectly, so there is still some chemical energy stored that is not converted to heat. With either a burner requiring fuel or an electric burner, not all of the heat will go into the water. Much of the heat is lost to the surrounding environment. Once again, for purposes of the calculation, we will assume we have a perfect system and no energy is lost in the conversion from chemical energy (fuel in a gas burner) or electrical energy (electric burner).

Thus, under the ideal system assumption, the energy required to heat the water is simply calculated as follows: heat energy equals specific heat capacity of the water multiplied by the mass of water multiplied by the change in temperature:

We will assume that the specific heat capacity of water is constant (even though in reality it varies with temperature), which is a pretty good assumption as it doesn’t change much in the temperature range we are looking at. First, using English units:

Since the definition of the British thermal unit was made for such problems, the calculation is fairly simple and the answer is 350 Btu’s.

Now, in SI units:

With the results from each calculation, we are ready to make some conclusions.

Conclusions

Hopefully, one of the first things you notice is that the energy involved in heating the pot of water (380,000 joules) is vastly greater than the energy involved in lifting up the bag of potatoes (110 joules). In fact, the potatoes would need to be lifted up to the countertop about 3,500 times to equal the amount of energy required to heat the pot of water. It sure is a good thing we have a burner to add that energy for us so we don’t have to add the energy mechanically. That would be a lot of lifting to get that water to boil!

Another item to notice is that we could immediately compare the values when we used SI units. The joule is an effective unit of energy when working with both mechanical energy and thermal energy. For the English units, however, we have two different units for the different kinds of energy. In order to compare the mechanical energy (80 foot-pounds) to the heat energy (350 Btu), you would need to know that there is a conversion factor of about 788 to convert foot-pounds to Btu’s. Thus, the 350 Btu’s is equivalent to 280,000 foot-pounds, which is a 3,500 times greater than 80 foot-pounds.

So, while using Btu’s or foot-pounds works just fine in some energy calculations, both of these units were made for specific kinds of energy and don’t easily relate to one another. I believe we are generally better off thinking about joules, an international standard that is easy to compare from one form of energy to another. The joule was designed to make sense as part of the overall SI system without getting messy and confusing like English units.

Another conclusion we can make is that while energy can be converted from one form to another, it is rarely very clean. There are always losses due to inefficiency. This is why experiments performed in the 1840’s to demonstrate the mechanical equivalent of heat (principally performed by James Prescott Joule) were so important and so revolutionary. The principle now known as conservation of energy (that energy is converted from one form to another such that energy is never lost in the process) was not obvious prior to those experiments. Energy is always “lost” in energy conversions, but it is never really lost in that it always goes somewhere.

Finally, as we already discovered, it takes a whole lot more energy to cook those potatoes than to get them up on the counter to prepare them. And not just by a little bit, but by a huge amount. This isn’t necessarily intuitive to us in daily life. But now you know the secret and are aware of the relatively enormous amount of energy required to change the temperature of water. So next time you are cooking on the stove, just think, that is a lot of energy going into that water!

Picture of potatoes in boiling water in a pot on the stove.