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Energy loss when boiling in an water experiment -- help please

  1. Dec 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi, as a part of my lab report I have to conduct this experiment : Fill a pot with tap water and boil it, determine then how much of the energy that the kitchen surface produced, actually went to the water itself. Consider the water having an initial temperature of 10 °C. In order to conduct the experiment I used an induction surface cook top. I filled a pot with 1litre of tap water and recorded the time it took to boil, which was 163 seconds. I then found out the power output of my cooking surface online which is 1.9 kW.

    2. Relevant equations
    E thermal = c *m* (ΔT)
    c = materials thermal constant , which for water is 4.18 kJ/kg°C (constant given by my book)
    ΔT= Difference in temperature

    P= E/t (Watt formula)

    3. The attempt at a solution
    First, I tried to find the energy produced by my surface during the time it took to boil the water:
    P=E/t , E=P*t , E= ( 1.9*10^3 W)* (163 s) = 309 700J .
    Then I tried to find the energy needed to turn a 1 liter of water from 10°C to 100°C. 1 liter of water has a mass of 1 kg

    E = c*m* ΔT = (4.18 * 10^3 ) * (1kg) * (100-10)= 376 200 j

    The energy expended by the cooking surface should not be smaller than the energy needed to cook the water. Where have I made my mistake?
    Sadly, i don't have access to a thermometer in order to measure the exact temperature of the water coming out of the tap. Also, I am using an induction surface to cook the water. Could it be that I am using a wrong formula in my calculations due to the fact that induction surfaces use electromagnetism in order to heat the pot directly?

    Do you have any ideas?
     
  2. jcsd
  3. Dec 22, 2015 #2

    Bystander

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    Think about it; how much air is dissolved in tap water?
     
  4. Dec 22, 2015 #3

    SteamKing

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    How do you know all of the water in the pot has reached a temperature of 100° C?

    If you don't have a thermometer to measure temperatures, how can you be sure of anything?

    Remember, the boiling point of water is also dependent on the ambient atmospheric pressure. Water doesn't boil at 100° C if you are located in La Paz, Bolivia (altitude 3650 m above sea level).

    Using tap water is also not recommended. There may be dissolved minerals, air, etc. which affect the boiling point.
     
  5. Dec 23, 2015 #4
    Hi, thank you for replying, I live in Stockholm and I searched online for the percentage of air in Stockholm water, but could not find any useful number, though it probably is lower than most places because the water in my area has high mineral content, which negatively affects its ability to absorb oxygen. I hope I understood your question correctly.
    Hi, thank you for replying as well. Unfortunately, I have no access to pure water or a thermometer that could do the job, but the assignment doesn't require access to those things either. I assumed the water had reached a temperature of approximately 100 °C because it had just started becoming steam when I finished counting the time. However, because of the high mineral content of Swedish water, it could be assumed that the water boiled at somewhat lower temperature. I don't think that the altitude was an important factor since the area I live in is only 28 meters above sea level.
     
  6. Dec 23, 2015 #5

    Bystander

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    NOT boiling.
     
  7. Dec 23, 2015 #6

    SteamKing

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    Water heated in a pot has convection currents set up in it by the heat which enters the bottom of the pot. The heat entering the bottom of the pot will create some bubbles of steam, which being lighter than liquid water, will rise to the surface. Just because you see some bubbles of steam, you should not infer that the entire contents of the pot are all at the same temperature, which is why scientists use instruments like thermometers to measure temperature rather than guessing.

    Also of consequence is the mineral content of tap water. A solution which contains dissolved minerals may not boil at 100° C but at a higher temperature, depending on the minerals in solution and their concentration:

    https://en.wikipedia.org/wiki/Boiling-point_elevation

    I don't know what the value of this 'experiment' is, but IMO, it seems to be pretty poorly designed. Even casual science should be more rigorous than this.
     
  8. Dec 23, 2015 #7
    Thank you both very much! I will redo the whole thing, but this time I will wait longer, until I can see that all the water is at a boiling point.
    I agree, but I don't think they intended it to be a proper experiment. The course I am taking is 1 month long distance course and covers the physics material of 1.5 years of high school. It is intended for either very bad students that failed the tests or immigrants that want to sit through the national examinations in order to get to university ( that's me). I think that the whole goal of the lab report is that the students show that they understand the concepts of energy and power, that they can do basic algebra and follow the expected form of a lab report (list material, correct referencing etc.). They do however expect us to write a list of all the things that could have contributed to us getting wrong measurements.
     
  9. Dec 24, 2015 #8

    CWatters

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    Your results are about 20% out but that doesn't allow for the efficiency of the induction hob or heating the pan so they could be more like 30-40% out..

    https://en.wikipedia.org/wiki/Induction_cooking
    That's quite a large error. I'm thinking the water from the tap was quite a bit warmer and or it wasn't close to being a uniform 100C at the end. I suppose the element could also dissipate more than it's rated power but that's not normally the case. Rating plates tend to overstate rather than under.
     
  10. Dec 24, 2015 #9

    Merlin3189

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    I am rather sceptical about the dissolved tapwater affecting the BP. As usual I'm no expert, but a quick search suggests 4 m mol/L would be very hard water and the highest solute concentration found in inland water (I would hesitate to call this "fresh water") by a USGS was about 125 m mol/L. This led me to estimate that BP could be elevated by 0.006 K for very hard water, or 0.2 K for the extreme sample. Since the OP requires to account for a discrepancy of about 16 K, even allowing for my quick rough estimates, I don't think solutes could do it.

    I would agree with CW that inaccurate estimates of temperature are more likely. Bystander is probably right, "not boiling".
    I also agree that the induction hob is unlikely to heat the water much more than calculated. I think the 1.9 kW is likely to be the input power and, although induction hobs are quite efficient, some is lost to heat in the electronics and coil (mine even has a small fan to cool the internals.)

    I am inclined to disagree with SteamKing about the value of the experiment. It has clearly provoked some serious thought by Stavrosnt to gain some insight into the process and calculations. What other objective could such an experiment have had? Even with purified water, a thermometer and barometer it would hardly be likely to enable a good estimate of the specific heat of water, or the efficiency of the hob for example. If it were a bad experiment, it would be because it was badly timed, so that the student still had too little confidence in his calculations to believe what they told him.
     
  11. Dec 24, 2015 #10

    OmCheeto

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    I concur.

    I just recreated the experiment, and the only thing I got at t = 163 seconds was a very wispy fog. (Temperature was about 43°C at that point)
    It took several hundred more seconds before my water actually started boiling.

    My vessel is aluminum, and weighs 750 grams. (for Al, c = 0.9 joules/(gm °C) )
    Tc = 16°C
    so
    ΔT = 27°C
    E thermal = c *m* (ΔT)
    1000 gm water --> 4.19 * 1000 * 27 = 113,000 joules
    750 gm Al --> 0.9 * 750 * 27 = 18,200 joules
    total = 131,000 joules

    time = 163
    joule/sec = watt
    watts = 131,000/163 = 800 watts

    I have no idea what my resistive burners are rated at, but I'd say 800 watts looks like a very good number, considering the myriad of other unmeasured factors involved.

    ps. I'm guessing it was at t ≈ 480 secs that things really started boiling
    so, 100-16 = 84°C
    4.19*1000*84=352,000 joules (water)
    0.9*750*84=56,700 joules (Al)
    total = 409,000 joules
    409000 joules / 480 seconds = 852 watts

    hmmm....
    burner resistance = 46.3 Ω
    stove voltage = 245 VAC
    (245 V)^2 / 46.3 Ω = 1300 watts (small burner)

    Much better than the two orders of magnitude that I'm usually off by, when experimenting. :redface:
     
  12. Dec 25, 2015 #11
    Thank you OmCheeto for taking the time to recreate the experiment!
    I have a question though: Is it safe to assume that the aluminum pot was the first thing to start being heated? If so, could we assume that by the time the water boiled the pot had surpassed the 100 °C mark and hence taken a bit more energy for itself? I am not talking big numbers here since the pot was dissipating the heat all that time, but could someone say that the result could have been affected by the pot reaching a higher temperature than 100°C during the time it took the water to boil?
     
  13. Dec 25, 2015 #12

    SteamKing

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    Yes, that is also a valid source for the difference in your numbers.
    Aluminum has a specific heat capacity of 0.9 J / g-°C, or 0.9 kJ / kg-°C, which means that changing the temperature of the pot 100° C, for example, requires the addition of 90 kJ of heat for each kg of the mass of the pot.
     
  14. Dec 25, 2015 #13

    OmCheeto

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    I doubt anyone would argue with this. [edit: Unless you count the burner itself. Mine weighs 190 grams. So I think we can ignore it, for the most part.]
    This seems like a very logical assumption.
    Very good conclusion, IMHO.

    And since you don't appear to have any temperature measuring devices, I went ahead and did another experiment, which involved measuring the temperature of my aluminum pot, with no water.

    I'm not quite sure if I'll have time to figure out what the data means, before I have to leave for a Christmas cruise with my friends at the river, but the graph looks very interesting.

    heating.and.cooling.an.empty.pot.on.the.stove.png
    x axis = time in seconds
    y axis = temperature in °C

    Notes:
    100°C was reached at time ≈ 70 seconds on the upswing, and ≈ 570 seconds on the downswing.
    Somewhere very close to time = 200 seconds, the burner was turned off.
    Random temperatures observed at ≈ 700 seconds:
    Handle of the pot: 18.6 °C
    stovetop: 55°C
    tile: 28.6°C

    dry.cooked.pot.thermal.experiment.2015.12.25.jpg

    Given the rise in temperature of the tile, my guess is that radiant heat loss, should not be discounted.
     
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