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Thermal Conductivity

  1. Jan 31, 2005 #1
    Hello,
    This is a homework problem for my Introductory Physics(no calculus) class but it's relatively simple so I opted to put it under high school. First I'll state the problem then my thoughts.

    A copper rod (k = 390) has a length of 1.5 m and a cross-sectional area of 4.00*10^-4 m^2. One end of the rod is in contact with boiling water and the other with a mixture of water and ice. What is the mass of ice per second that melts? Assume that no heat is lost through the side surface of the rod.

    This seems simple enough. Obviously
    Q=(390*(4*10^-4)*DT*t)/1.5
    DT is where I am coming across a problem. Am I supposed to assume that it is 100-0? Doing so gives
    Q=10.4*t J*s
    My idea is that in order for the ice to melt it has to have at least m*L (L=33.5*10^4) so I set 10.4*t=m*L
    and it follows m/t=10.4/(33.5*10^-4) which is approximately 3.1045*10^-5 kg/s

    I would greatly appreciate it if any could help me with this. I can't seem to find much information on this type of problem and I've looked to other books. They seem to have this problem, but no solution.
    Anyway, I'm not looking for an answer just trying to see if my logic is correct. Thank you in advance
     
  2. jcsd
  3. Jan 31, 2005 #2

    Andrew Mason

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    Since the question asks for the rate of ice mass melting/second, you are interested in finding Q/t. This is proportional to the thermal conductivity, area and temperature difference and inversely proportional to the length of the conducting body (rod). You have to use the entire temperature difference, which is 100 degrees. That gives you Q/t of 10.4 J/sec. as you have found, which looks correct.

    As you seem to realize, you have to figure out how much heat flow is required to melt one gram or kg. of ice per second. The latent heat of water is 334 kJ./kg. or 334 J/g.

    The rest is just plugging in the numbers.

    AM
     
  4. Jan 31, 2005 #3
    Thank you for your reply. I understand what you are saying I just can't tell if I've set it up correctly or not. There is 10.4 J/s or W going to the mixture, I know it takes 33.5*10^4 J to melt 1 kg of ice, but it is the time factor that is confusing me. Is this Q/t=(mL)/t okay, because if I don't put the time on the other side I get something like kg = kg/s
    Anyway, I still come up with 3.1045*10^-5 kg/s approximately.
     
  5. Jan 31, 2005 #4

    Andrew Mason

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    That looks right. It takes about 32 seconds to melt 1 gram.

    AM
     
  6. Jan 31, 2005 #5
    Alright, thank you again.
     
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