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Thermodynamics speed of a weight bar through ice

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data

    A steel bar of rectangular cross-section, with height a and width b (into the paper), is placed on a block of ice with its ends extending a little beyond the ice, (ice has a length c). A weight of mass m is hung from each end of the bar. The entire system is at 0 degrees celcius.

    As a result of the pressure exerted by the bar, the ice melts beneath the bar and refreezes above the bar. Heat is therefore liberated above the bar, conducted through the metal, and then absorbed by the ice beneath the bar. Given the latent heat of fusion per kilogram of ice (L), the density of ice ρi, the density of water ρw, the thermal conductivity of steel (Kappa) which relates the heat q crossing a unit area per unit time to the temperature gradient (dT/dz) in a direction perpendicular to the plane through the relationship q=-(kappa)(dT/dz), the temp T(=0 degrees celcius) of the ice, the accelertation due to gravity g, the mass m, and the dimensions a, b, and c, where c is the length of the block of ice, show that the speed v with which the bar sinks through the ice is


    v= [(2mg(kappa)T)/(abcL2ρi)]*[(1/ρi)-(1/ρw)]


    2. Relevant equations

    A=bc

    dP/dT = L/TΔV ~ ΔP/ΔT

    dq = -(kappa)(ΔT/Δz) ( my prof wrote it as q with a dot over top so i assume it's some sort of dq)

    (dQfusion/dt) = (dM/dt)(L/A)

    3. The attempt at a solution

    Everytime i look at it, i don't see a constant velocity since the pressure should be increasing therefore the temperature below the bar would be increasing. anyways

    I thought to try and relate the equations with ΔT to get rid of the ΔT

    so,

    dq= -(kappa)(ΔT/Δz) = -(kappa)([ΔPTΔV]/LΔz)

    then for the dQfusion i thought to multiply through by dt and equate the above equation dq to dQfusion

    so

    dz= -(kappa)*([ΔP*T*ΔV*A]/[dM*L2)


    then i thought that ΔP has to be 2(dM)g/A or 2mg/A (since it's in the top of the answer)

    ΔV = Adz


    that's all i have and i don't think it's right.

    any help would be appreciated
     
  2. jcsd
  3. Dec 6, 2011 #2
    guess i could answer my own question, for anyone who has a similar one.

    first of all Δv is the molar density

    and dq is supossed to be (dq/dt) which is equal to the dQfusion/dt

    so

    dq/dt=dQfusion/dt

    => -(kappa)(ΔT/Δz) = (dM/dt)(L/A)

    and (ΔT/Δz)=(ΔT/a) since Δz=a-0

    move L/A over

    (dM/dt)= -(kappa)(ΔTA/aL)

    next you want to solve for ΔT from the dP/dT

    ΔT = (ΔPTΔv)/L

    where ΔP = -2mg/bc

    so then we have

    dM/dt = [2mg(kappa)TΔvA]/[L2abc]

    now you need to related the change in mass with respect to time with the change in position of the bar with respect to time

    dM/dt = ρibc(dy/dt)

    therefore

    dy/dt = (dM/dt)(1/(ρibc))

    then

    dy/dt = [2mgT(kappa)Δv]/[L2abcρi]

    the disappears because A=bc

    molar volume is equal to 1/ρ

    which gives the final answer of


    v= [(2mg(kappa)T)/(abcL2ρi)]*[(1/ρi)-(1/ρw)] = dy/dt
     
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