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  • Calculating energy loss

    Jennifer Murphy

    Jennifer Murphy

    @jennifer-murphy-CyNi39
    Updated: Oct 26, 2024
    Views: 1.3K
    Suppose I have a 14 oz insulated thermos. I fill it with water heated to 145°F. After an hour, I measure the water temperature as 135°F. How do I calculate the energy loss?

    To calculate the number of calories lost, I converted the temperatures to °C, found the difference, and multiplied by the mass converted to grams.
    • 135°F = 57.2°C
    • 145°F = 62.8°C (Corrected, originally posted as 145°F = 82.8°C)
    • 62.8-57.2 = 5.6°C (Corrected, originally posted as 82.8-57.2)
    • 14 oz = 396.9 grams
    Then, calories = 396.9 x 5.6 = 2,205

    Is that right?

    This loss happened over an hour, so the loss/minute would be 2205 / 60 = 36.7

    So if I wanted to maintain a constant temperature, I would need 36.7 calories/minute?

    Correct?

    Comments?

    Thanks
    0
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Replies

  • Ramani Aswath

    Member • Mar 14, 2015

    The conversion and the arithmetic seems bit wonky. However the difference 5.6C is correct as is the rest.
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  • Jennifer Murphy

    Member • Mar 14, 2015

    A.V.Ramani
    The conversion and the arithmetic seems bit wonky.
    Wonky? How so? This is an engineering forum, no? 😀

    A.V.Ramani
    However the difference 5.6C is correct as is the rest.
    So then what would it take to put 37 cal/min into the system? Could it be done with the power from a USB port attached to a computer?
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  • Ramani Aswath

    Member • Mar 14, 2015

    2.25 Kcals approximately equal to 2.6 W, well within the capacity of the USB port. Actually an induction heater coffee warmer working off an USB port is already in the market.
    #-Link-Snipped-#
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  • Jennifer Murphy

    Member • Mar 15, 2015

    A.V.Ramani
    2.25 Kcals approximately equal to 2.6 W, well within the capacity of the USB port.
    Thanks
    Actually an induction heater coffee warmer working off an USB port is already in the market.
    #-Link-Snipped-#
    This is not an induction heater. The mug isn't included. It works with any coffee mug, so there can't be any heating coils in the mug. It is just a regular mug heater that happens to be powered by the USB port, and apparently not a very good one at that.

    Here's another listing with a comment stating that the mug is not included:
    <a href="https://www.amazon.com/TOOGOO-USB-Warmer-Heater-Port/dp/B00CJ3JFZM/ref=sr_1_4?ie=UTF8&qid=1426419339&sr=8-4&keywords=usb+coffee+mug+warmers" target="_blank" rel="nofollow noopener noreferrer">Amazon.com</a>

    And here's another listing showing it without the mug:
    <a href="https://www.amazon.com/Laptop-Buddy-Powered-Beverage-Warmer/dp/B003TNWIB4/ref=sr_1_8?ie=UTF8&qid=1426419339&sr=8-8&keywords=usb+coffee+mug+warmers" target="_blank" rel="nofollow noopener noreferrer">Amazon.com</a>
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  • Ramani Aswath

    Member • Mar 15, 2015

    Jennifer Murphy
    This is not an induction heater.
    Maybe this one is not.
    However, one induction cup warmer powered through USB was developed as a concept proving exercise. Shall put that up if I locate the item.
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  • Jennifer Murphy

    Member • Mar 15, 2015

    I did find a couple of devices that claim to use induction heating.

    This one appears to actually use induction technology.

    <a href="https://www.gadgetify.com/revollexs-induction-coffee-cup-warmer-heat-your-drink-not-your-cup/" target="_blank" rel="nofollow noopener noreferrer">Revollex's Induction Coffee Cup Warmer: Heat Your Drink Not Your Cup</a>

    It requires that a metal disk be placed in the bottom of the mug. If the claims are accurate, it works well, but it has a number of problems.

    Here's the Indiegogo listing from back in 2013:

    <a href="https://www.indiegogo.com/projects/paragon-induction-cooktop#cookware" target="_blank" rel="nofollow noopener noreferrer">Paragon Induction Cooktop | Indiegogo</a>

    They only raised $270 of the $320,000 goal.


    This one claims to use induction, but it looks exactly like the one you posted.

    #-Link-Snipped-#
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  • Jennifer Murphy

    Member • Mar 15, 2015

    I think there may be two similar, but distinct, concepts here that might be getting mixed up. The two concepts are "heat induction" (or "induction heating") and "magnetic induction". Here's my understanding:

    Magnetic induction is the effect discovered by Faraday and others in the mid 1800s in which an electric current is induced when a conductor is passed through a magnetic field or, conversely, when a stationary conductor is exposed to an oscillating magnetic field. Correct?

    Induction heating is the process of heating an electrical conductor (usually a metal) by electromagnetic induction, where eddy currents (also called Foucault currents) are generated within the metal and resistance leads to Joule heating of the metal. Correct?
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  • lal

    Member • Mar 15, 2015

    Could my calculator be misbehaving? I got 82.8-57.2 = 30.1°C.
    Or am I missing something?

    By the way, what you said about Induction heating is right.
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  • Jennifer Murphy

    Member • Mar 15, 2015

    lal
    Could my calculator be misbehaving? I got 82.8-57.2 = 30.1°C.
    Or am I missing something?
    Good catch. Typo. Should have been 62.8-57.2 = 5.6°C.
    I fixed it.

    How come you didn't catch the first error in the conversion of 145°F? 😉

    By the way, what you said about Induction heating is right.
    Thanks
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  • Ramani Aswath

    Member • Mar 15, 2015

    Jennifer Murphy
    Correct?
    On the nose, correct.
    Eddy current losses appear as heat in the metal, which is in essence a shorted secondary of a transformer. Such a device has an efficiency nearly matching an immersion heater.
    The experimental device I posted about was such a device running off a USB port. It does not heat the liquid from cold but just matches the heat loss, which in your original post came to about 2.5 W, the limit for USB outputs now. However, it did not reach the market. The Revolex unit runs off mains and has a much higher wattage.

    (As an aside, the difference in temperature is 10 F, which when divided by 1.8 is 5.5555...recurring or 5.6 C, which is what I did.)
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  • Shashank Moghe

    Member • Mar 16, 2015

    mass*Cp*dT

    and NOT

    mass*dT
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  • Shashank Moghe

    Member • Mar 16, 2015

    But then, the Cp of water is 1 cal/gram-C. So you are good omitting the Cp factor there.
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  • lal

    Member • Mar 16, 2015

    Guess I'm more comfortable with Degrees than Fahrenheit, hence I didn't take the pain to convert those 😁

    I was thinking why wouldn't USB cup warmers be pleasing most cutomers. Temperature drop greatly depends on the external temperature too. A normal USB 2.0 port can provide a maximum of only 500mA current at 5V (and that too if you are not connecting from a HUB). That translates to 2.5 watts ignoring any losses, which might not be enough to keep a cup of coffee 'warm' in colder environment.

    Things get better with USB 3.0, where the maximum current capacity is 900mA. That is 4.5W power output. Fair enough! Alternately, USB 3.0 enables devices to draw around 1.5A in a special charging mode in which data transfer will be disabled. Things get better, 7.5W power output!

    If you have to go the induction way, that would complicate things further. A special circuit to oscillate the otherwise stable DC 5V USB output at high frequency Alternating supply for inducing emf and hence eddy current in the cup/pan/mug's metal base; and that too at such a low potential.
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  • Jennifer Murphy

    Member • Mar 16, 2015

    lal
    Guess I'm more comfortable with Degrees than Fahrenheit, hence I didn't take the pain to convert those 😁
    And I just realized that my digital thermometer has a C/F setting, so I could have gotten the readings in C in the first place.

    I was thinking why wouldn't USB cup warmers be pleasing most customers.
    I am quite sure they would be if they worked. But they don't. The ones that are essentially a hot plate are a disaster. Most cups have a thin rim on the bottom, so they make very little contact with the hot plate. Most of the heat is lost to the environment. And most of these are running off of A/C so they have plenty or power. But in order to heat the cup, they would have to be too hot to be safe. And they would burn the coffee, not to mention fingers.

    Temperature drop greatly depends on the external temperature too.
    Yes, but I'm only interested in an office environment, so about 21°C.

    I'm not wedded to USB power. I doubt many people would use a USB cup warmer on a laptop that was not plugged into a wall outlet. I would think the cup warmer would drain the battery fairly quickly. So there should be an outlet reasonably handy.

    If you have to go the induction way, that would complicate things further. A special circuit to oscillate the otherwise stable DC 5V USB output at high frequency Alternating supply for inducing emf and hence eddy current in the cup/pan/mug's metal base; and that too at such a low potential.
    Well, maybe USB won't work. As I said, I don't care about that.

    My real objective in posting was to make sure I was calculating heat loss correctly. I like to keep a thermos of a hot beverage (usually tea) by my desk. I have been frustrated for a long time by having it too hot or too cold. I dream of a "Goldlilocks Thermos" that keeps the liquid at just the right temperature.

    According to my measurements, the heat loss in a good thermos is fairly small over a few hours. It seems like someone could design a heated thermos that would keep the contents at a more or less constant temperature.

    I was attracted to the idea of induction heating, because that would allow for a sealed thermos bottle that could be used without the base. But one with a copper connector would be OK, too. I just worry that it would get corroded by spilled coffee.
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  • Jeffrey Arulraj

    Member • Mar 16, 2015

    Generating an alternating field using an USB I seriously doubt it mate.

    Avoid it if needed, Cos you will need to protect your USB from alternating voltage reflected from the heater. Since the load on the heater is varying with the load( Coffee cup size) designing an Isolation circuit is hard.

    Make sure you use a really worn out computer if you plan on testing this out. I am not sure whether this can work
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  • Jennifer Murphy

    Member • Mar 16, 2015

    Jeffrey Samuel
    Generating an alternating field using an USB I seriously doubt it mate.

    Avoid it if needed, Cos you will need to protect your USB from alternating voltage reflected from the heater. Since the load on the heater is varying with the load( Coffee cup size) designing an Isolation circuit is hard.
    OK. USB won't work.

    Make sure you use a really worn out computer if you plan on testing this out. I am not sure whether this can work
    I'm not going to be testing it. I don't have the skills or the means. I'm a software engineer. We are not even allowed to use screwdrivers.

    I'm just trying determine whether it's feasible and maybe get a handle on how to do it. Then I'll see if I can persuade someone to actually build it. All I want is a nice hot drink on my desk all day long.
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  • Jeffrey Arulraj

    Member • Mar 16, 2015

    Try out #-Link-Snipped-# idea that will work for sure. But nice thread and observation I would say
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  • lal

    Member • Mar 16, 2015

    I wouldn't doubt it remains a threat to the PC. We, the humans, have been using mere 12V or even 6V batteries safely to create alternating output [inverters]. a properly designed circuit should provide enough isolation.

    But, the things we must be concerned should be the 'EMI/EMC - Electromagnetic interferance/ Electromagnetic compatibility' of the device. At such high frequency, it would be a good source for noise. A well designed shielded connection and filtering in all possible means would be unavoidable to make sure the computer or other electronic equipments around are not affected.

    ---------------------------------------------------

    Anyway, your requirement is easy to achieve with a simple closed loop system. A thermal sensor inside the thermos, a heating method (be it inductive heating or resistive) and a controller (quoting Raspberry Pi as an example controller) should help and much easier if you can supply from the mains.

    Set the required temperature and sit back. Any loss will be automatically compensated by actuating the heating by the controller. Though you might want to keep the lid open in case the temperature is above required limit 😁
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