1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximum mass of a metal you can melt; Heat of Transformation

  1. Jul 19, 2015 #1

    KAC

    User Avatar

    1. The problem statement, all variables and given/known data
    What is the maximum mass of lead you could melt with 2000 J of heat, starting from 25 ∘C ? Lead melts at 328∘C , its specific heat is 128 J/(kg⋅K) , and its heat of fusion is 2.5×10^4 J/kg .

    2. Relevant equations
    Need to find both the mass in the heat of transformation (Q = MLf) and the mass that resonates with the outcome of Q=McT.

    3. The attempt at a solution
    With my first attempt, I believed that once I worked out both equations, I would subtract the product of Q=MLf from Q=McT, but that was incorrect. My work is as follows:

    Q=McT
    M=cT/Q
    M= (128 J/kg*K * (328-25)K)/2000J = (128)(303)/2000; M = 19.39 kg.
    (Difference in two temperatures of Celsius is equal to the difference in Kelvin, so I made the substitution from C to K here)

    Q=MLf
    M= Lf/Q; (2.5*10^4 J/kg)/(2000J) = 12.5kg

    Because 12.5kg is the maximum mass that this heat can transform altogether, and 19.39kg is the mass that can be raised in temperature from the first equation, I subtracted one from the other. 19.39kg-12.5kg=6.89kg, but this is incorrect. Can anyone steer me in the correct direction for this problem?
     
  2. jcsd
  3. Jul 19, 2015 #2

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You're not asked to do one or the other, raise to fusion temperature or fuse metal at fusion temperature, but to raise it to fusion temperature AND fuse it.
     
  4. Jul 19, 2015 #3

    KAC

    User Avatar

    So for this, how could I set it up? Should I add the two products together or set the two equations equal to each other such as cT/Q = Lf/Q?
     
  5. Jul 19, 2015 #4

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    This is PF. We try to get you to solve the problem. What do you think should be the next step?
     
  6. Jul 19, 2015 #5

    KAC

    User Avatar

    I am trying to solve the problem.. But I feel like I'm going in the completely wrong direction with the way I've set up my problem solving, so I would appreciate the assistance of pointing out one part of my work that I might have calculated wrong or an equation that I shouldn't be using at all.
    I need the first equation Q=McT because I need to get the mass that correlates with how much the temperature rises before phase change, at which point there would be no more temperature change until the entire solid became a liquid. I used the second equation, Q = MLf because Lf is the heat of fusion which is the temperature at which the metal will be once it fully reaches a liquid state. I used the same Q in both problems because it is the amount of heat that is being used. I was temporarily thinking I could add the masses together, but that doesn't make sense because how can the mass be greater than what it could be for the solid to begin transforming into a liquid with the heat that is given?
    So, since that is out of the question, one last thing I can think to do is to solve Q = McT and use the M= 19.39kg and plug this into Q=MLf. Now if I do this I get the product 484750 that the heat would need to be to perform this transition from solid to liquid form, but I'm not even sure where to use this or if it even makes sense. I'm really under the impression that this problem implies that the heat applied remains constant throughout the transition, so Q would be 2000J for both problems and I would really only need to solve for the mass.
    So if I try to set the equations equal to one another to solve for the mass, I would be setting the equation up as M = Lf/Q & M = cT/Q, so Lf/Q = cT/Q; 12.5 kg*K = 19.39 kg; but this doesn't work out and all of the units cancel except for K, which I'm not looking for at all?
     
    Last edited: Jul 19, 2015
  7. Jul 20, 2015 #6

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You are given 2000 joules heat, and you have TWO things to do with that 2000 joules, increase the temperature of a mass of metal to its fusion point, and fuse that mass.
     
  8. Jul 20, 2015 #7

    KAC

    User Avatar

    Okay, so what I am thinking now, since 2000J is only given to us once, is that Q = McT + MLf. So,

    Q = McT + MLf; Q = 2M(cT + Lf); 2M = Q/(cT + Lf). When calculating this out I get:

    2M = 2000J/[(128J/kg*K * (328-25)K) + 2.5 * 10^4 J/kg]; 2M = 0.03136kg; 0.03136/2 = 0.0157 kg or 15.6g would be the mass, but this also turns out to be wrong.

    I really thought that the first direction I was going in was correct, because Q = Q(Heat to raise 1kg by 1K) + Q(Heat for phase change) since I have to use both equations separately. So if I do that and use what I calculate for both changes, I would get an addition of 19.39kg+12.5kg, which equals to be 31.9kg. But I don't believe this to be correct because I don't understand how the total mass could be more than what is available at either of the two changes. From this point, another thought I have is possibly dividing this 31.9kg by 2 since there is only 2000J for both changes. If 31.9kg/2, I get the result 15.9kg. Does this sound like the correct path to working it out?
     
    Last edited: Jul 20, 2015
  9. Jul 20, 2015 #8

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Two?
     
  10. Jul 20, 2015 #9

    KAC

    User Avatar

    The way that I computed it, is that since both equations had M in the formula, I would then have 2M after adding the formulas together, and after completing the work I'd need to divide my answer by two to account for this.. But now I see where that doesn't make sense! Since I'm taking out a common factor, I wouldn't have two M, I'd still have one... So let me try it by only having 1M. In this case:

    Q = McT + MLf; Q = M(cT + Lf); M = Q/(cT + Lf); M = 2000J/[(2.5*10^4 J/kg * (328-25)K) + 128J/kg*K]; M = 2.64*10^-4kg, or 0.26g.
    =
    This also turned out to be incorrect, and the correct answer was 31.4g. So, what I am thinking, is that 19.36kg (From Q=McT) + 12.5kg (From Q=MLf) = 31.4 kg. However, the answer that was correct was in the units of g, so I'm completely lost on how this change occurred! When I compute the answer by adding the two products together, is it simply by chance that I'm getting an answer that is almost correct, and I'm really using the wrong math work?
     
  11. Jul 20, 2015 #10

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Heat of fusion x temperature change? You cannot just throw numbers into an equation and expect the calculator to intuit what's to be done with them. If it is necessary to proceed one step at a time, do so.
     
  12. Jul 20, 2015 #11

    KAC

    User Avatar

    THANK YOU for pointing that out. I think I'm crunching numbers so much I really just am throwing them together.. but yes, I should have plugged 128 for c; I mistakenly multiplied the change in temperature by the heat of fusion and added that to the specific heat when I should have done it the other way around. This is my mistake fully. So when I compute that, I DO come out with 0.0319kg, which converts to 31.9g. With this, I am going to rework the problem; please do critique my work and the steps I take if you have the available time!

    Since the problem calls for heating the material to phase change, then through the phase change, I use Q=McT as the formula for the specific heat that is required to raise the temperature of this specific material. I use Q=MLf as the formula for the heat required for the material to change from its solid state to liquid state. Q = Q(Temperature) + Q(Heat of Fusion, change from solid to liquid state).
    Specific Heat the material requires:
    Q = McT
    M = cT/Q
    M= [(128J/kg*K)*(328-25)K]/2000J
    The units of K & J cancel out, so I am left with M=19.39kg.

    Phase Change Heat Requirement:
    Q=MLf
    M=Lf/Q
    M=(2.5*10^4 J/kg)/2000J
    Units of J cancel out, so the product becomes M=12.5kg

    Finally, Q = Q(specific heat) + Q(heat of fusion), therefore M = M(specific heat) + M(heat of fusion), so M = 19.39kg + 12.5kg = 31.9kg is the maximum mass of lead that can be put through a phase change with this amount of heat.

    Thank you so very much for your help, Bystander!
     
  13. Jul 20, 2015 #12

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Might look better solving once, with a single value for Q (2000), for a single mass just as you did in #9;
    What you've done in #11 is making my head hurt --- it might be correct with a little more subscripting, but as it stands, it looks as though you've solved for two different masses, one to be heated, and another to be melted --- some people are going to break out in hives at the thought.
     
  14. Jul 20, 2015 #13

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Try looking at this problem algebraically.

    Assume that the mass of lead which can be fused with 2000 J of heat is x kilograms.

    Write an equation which tells you how many joules of heat it takes to raise x amount of lead from 25 C to the MP of lead.

    Write an equation which tells you how many joules of heat it takes to fuse x amount of lead.

    The sum of the heats required by two equations must equal 2000 J.

    Solve for x.
     
  15. Jul 20, 2015 #14

    KAC

    User Avatar

    So then I should set it up with the same equation Q = Q(specific heat) + Q(phase change)?:
    Q = McT + MLf
    Q = M(cT + Lf)
    M = Q/(cT + Lf)
    M = 2000J/(128J/kg*K * 303K) + 2.5*10^4 J/kg

    And with this work I did get the correct answer, M = 0.0314 kg; M = 31.4 g. Thanks again, this will help on my upcoming exam.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted