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Heating water in a kettle (time, boil out time, work done)

  1. Sep 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A domestic kettle is marked 250 V, 2.3 kW and the manufacturer claims that it will heat a pint of water to boiling point in 94 s.

    (a) Test this claim by calculation and state any simplifying assumptions you make.
    (b) If the kettle is left switched on after it boils, how long will it take to boil away half a pint of water measured from when it first boils?
    (c) Estimate the work done against an atmospheric pressure of 100 kPa when 1 cm3 of water evaporates at 100 °C, producing 1600 cm3 of steam. Express this as a percentage of the total energy required to evaporate 1 cm3 of water at 100 °C.

    • Specific heat capacity of water = 4.2 * 103 J kg-1 K-1
    • Specific latent heat of vaporisation of water = 2.3 * 106 J kg-1
    • Density of water = 1.0 g cm-3
    • 1 pint = 570 cm3

    Answers: (c) 1.6 * 102 J, 7.0 %

    2. The attempt at a solution
    (a) We need to find Δθ in ΔQ = m c Δθ. Q is V I t, where I = P / V = 2300 / 250 = 9.2 A.

    Then find Q = 250 * 9.2 * 94 = 216 200 J. m is 1 g cm-3 * 570 cm3 = 570 g. 570 g / 1000 = 0.57 kg. Substitute for Δθ = ΔQ / mc = 216 200 / (0.57 * 4.2 * 103) = 90.3 °C.

    Water boils at 100 °C, so the manufacturer is wrong. In 94 s we'll only get to 90.3 °C.

    (b) The mass of water that will be in the kettle at 100 °C: m = ΔQ / cΔθ = 216 200 / (100 * 4.2 * 103) = 0.51 kg.

    Find ΔQ with half of that mass: ΔQ = 0.26 * 4.2 * 103 * 100 = 108 100 J.

    Find time: t = Q / VI = 108 100 / 250 * 9.2 = 47 seconds. That is half the claimed boiling time. Maybe it was possible to just divide the boiling time (94 s) by two? Since at boiling time half of the water will boil out in half the time (same as a third in a third and 1/4 in 1/4 etc.).

    I didn't start (c) since I am not sure whether both (a) and (b) are correct. I also worked on this problem some while ago and in (a) I had a different mass = 0.473 kg and I got the temperature to be equal to 109.35 °C (so the manufacturer is right). But I can't remember at all how did I derive that number, maybe it is wrong and this is correct: m is 1 g cm-3 * 570 cm3 = 570 g. 570 g / 1000 = 0.57 kg.

    Any thoughts on (a) and (b) please?
     
  2. jcsd
  3. Sep 22, 2016 #2

    kuruman

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    (a) looks correct although you could have gotten the heat going in by using Δθ = P t directly. The problem is asking you to list your assumptions. What assumptions are you using to get your answer?
    (b) I would find the mass of the water differently. What is the mass of one pint of water? Use that instead.
     
  4. Sep 22, 2016 #3
    You meant ΔQ and not Δθ? Since P t is 216 200, same as Q. In terms of assumptions I didn't make any. Water boils at 100 degrees, so if the manufacturer is right then the temperature in 94 seconds will be the same or higher than 100. If wrong -- less.

    1 pint = 570 cm3
    density of water = 1.0 g cm-3

    570 * 1 = 570 g. In kg: 570 / 1000 = 0.57 kg.
     
  5. Sep 22, 2016 #4

    kuruman

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    Yes, I meant ΔQ, sorry for the confusion.
    Where did you get this number? Google "pint to cc conversion" and see what you get.

    On edit: Which side of the Atlantic are you on? One UK pint is 570 cc as you said, but 1 US pint is 470 cc. More importantly, which side of the Atlantic is the manufacturer of the heater who made this claim? Here is an assumption to add to the list.
     
    Last edited: Sep 22, 2016
  6. Sep 22, 2016 #5
    This number is given in the problem (along with others), so I did not look it up.
    The book is published in the UK, so I guess that's the reason for 570 cm3. Personally I am from Russia and we don't use pints, litres like the rest of Continental Europe.

    I think that that's the reason why in my previous tries I had the mass as 0.473 kg. I probably googled what one pint is and got 470 cm3. And using this number the temperature is 109 °C and the manufacturer is then correct. So we could assume that if the manufacturer is from the US, then the timing is true. And if the manufacturer is from the UK, then it's is wrong. If not your tip I would've never though of that actually, since a person from UK or USA is required to know these differences :), or at least someone who is familiar with the UK / USA differences.
     
  7. Sep 22, 2016 #6

    kuruman

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    OK, so one assumption is that the amount of water is 1 UK pint which has a mass of 0.57 kg. That's what you say in part (a). However, in part (b) you say that the mass is 0.510 kg. You need to fix that to be consistent. Also, it looks like you misunderstood what (b) is asking you to find. The 94 s of part (a) is irrelevant here. Here is the picture: You have 1 pint of boiling water. As soon as the water starts boiling, you start a timer. What does the timer read when half a pint of water has boiled away?

    Can you think of the assumptions that go into calculations of this kind other than the US/UK pint difference?
     
  8. Sep 22, 2016 #7

    gneill

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    I don't see where you've made any assumption about the starting temperature of the water. You've calculated a Δθ, but ffrom what initial temperature is the 'Δ'? Does it matter for your answer to (a)?
     
  9. Sep 22, 2016 #8
    Regarding (a):
    We're assuming that we are in the UK, since we have 1 pint = 570 cm3. We need to find to what temperature will the kettle boil the water in 94 seconds. Let's assume that the initial temperature is 0 °C.

    ΔQ = mcΔθ, where Δθ = ?
    To find ΔQ we use: ΔQ = Pt = 2300 * 94 = 216 200 J.
    Mass of the water is: 570 cm3 * water density 1 g cm-3 = 570 g or 0.57 kg.
    So Δθ = 216 200 / (0.57 * 4.2 * 103) = 90.3 °C

    The boiling temperature of water is 100 °C and therefore it is either a UK manufacturer is lying or the manufacturer is from the US and the situation in the US is different.

    I think this should be it in (a)?

    Update: regarding (b): ΔQ = 0.285 [half 1 pint = 0.57 kg] * 4.2 * 103 * 100 [boiling temp.] = 119 700 J. t = Q / t = 119 700 / 2300 = 52 s. So it will take 52 seconds to boil half the water.
     
  10. Sep 22, 2016 #9

    gneill

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    Or the manufacturer will assume that the water starts at room temperature so that his claim looks better and he can claim it's a more typical situation...

    edit: Did a quick google search on "average tap water temperature uk" and found out that it is about 7C, but varies with the season. :smile:
     
  11. Sep 22, 2016 #10

    kuruman

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    Here some more questions to consider
    Is the boiling temperature of water 100 oC everywhere on the surface of the Earth?
    Does all the power from the heater go into boiling water?
     
  12. Sep 22, 2016 #11
    That's also a point :).

    Yes, it also depends on the height.
    Probably part of it is lost somewhere in powering the buttons etc. So we consider that 100 % of power is used to heat water.

    What about (b)? ΔQ = 0.285 [half 1 pint = 0.57 kg] * 4.2 * 103 * 100 [boiling temp.] = 119 700 J. t = Q / t = 119 700 / 2300 = 52 s. So it will take 52 seconds to boil half the water. Is this correct?
     
    Last edited: Sep 22, 2016
  13. Sep 22, 2016 #12

    haruspex

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    I think you have misread the question. For this part, assume the water is already at boiling point. How long does it take for half of the water to boil away into vapour?
     
    Last edited by a moderator: May 8, 2017
  14. Sep 23, 2016 #13
    Hm, should it be then: ΔQ = ml = (0.57 / 2) * 2.3 * 106 = 655 500 J → t = Q / VI = 655 500 / (250 * 9.2) = 285 s? Isn't it too long?
     
  15. Sep 23, 2016 #14

    kuruman

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    I don't think so. It's less than 5 minutes.
     
  16. Sep 23, 2016 #15
    Why 5 minutes?
     
  17. Sep 23, 2016 #16
    Regarding (c) I think it should be: Work done W = pV (pressure * volume). p = 100 000 Pa and V = 1600 cm3 or 1.6 * 10-3 m3. So W = 100 000 * 1.6 * 10-3 = 160 J. This should be correct.

    "Express this as a percentage of the total energy required to evaporate 1 cm3 of water at 100 °C." for this one we have ΔQ = ml where m = 1 cm3 * 1 g cm-3 = 1 g or 1 * 10-3 kg. So Q = 10-3 * 2.3 * 106 = 2300 J. (160 / 2300) * 100 = 7 %.
     
  18. Sep 23, 2016 #17

    haruspex

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    Do you cook? Have you ever had to boil a sauce to reduce its volume? Seems to take forever.
     
  19. Sep 23, 2016 #18
    That's indeed so :).

    Alright, so (a) and (b) should be good. And (c)? I think it should be correct.
     
  20. Sep 23, 2016 #19

    kuruman

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    Five minutes is 300 s. Your answer is 285 s, less than 5 min.
    It looks OK.
     
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