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Thermo formula required.

  1. Apr 12, 2012 #1
    Can any one direct me to the correct Thermo formula to solve the following problem.
    What is the time for water to heat up from 25'c to 75'c when placed in a 1000'c oven?
  2. jcsd
  3. Apr 12, 2012 #2
    Fourier's law would govern this.

    The water wouldn't heat evenly, and if you want to know the time before the coldest part of the volume reaches 75 (by which time the edges will be hotter), you need to do some maths which factors in (a) the thermal conductance of the water and the container and (b) the shape of the water and thickness of the container.
  4. Apr 12, 2012 #3
    Tnx Mikey

    The furier law give me the heat transfer, but what is the equation to find the time for the water to heat up given the heat transferred?
    I have water running trough a copper pipe, entering the pipe at 25°c, the pipe is placed in a 1000°c oven, I'm trying to found out what should be the length of the pipe so that the water come out of the pipe at 75°c.
    I understand that the water in the pipe center will be colder then the water near the edges, but I need the average temperature of the water.

  5. Apr 12, 2012 #4


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    If the pipe is reasonably long in relation to diameter, I wouldn't worry too much about unequal heating.
    Pipe material conductivity = k
    Inner radius = r
    Outer radius = R
    Water temp at x from start = T(x)
    Oven temp = H
    Specific heat of water/unit vol = s
    velocity of water = V
    A = π.r^2
    A full analysis gets messy because there will be heat flow along the pipe (back towards point of entry into oven) as well as radially, but let's ignore that.
    Taking a slice through the pipe thickness dx, the radial heat flow depends on the temperature difference and the inner and outer radii:
    F(x) = dx.2π.k.(H-T(x))/ln(R/r)
    In time dt, heat entering dx of the pipe = F.dt
    In that time, a volume of water V.A.dt passes any given point.
    So in distance dx the water warms by dT = F/V.A.s
    dT/dx = 2π.k.(H-T(x))/(ln(R/r).V.A.s)
    ln((H-T)/c) = -2π.k.x/(ln(R/r).V.A.s), where c = H - T(0).
    If the pipe is length L, water emerges at temp T(L) :
    ln((H-T(L))/(H-T(0))) = -2π.k.L/(ln(R/r).V.A.s)

    Check the math, plug in the constants and solve for L.
  6. Apr 15, 2012 #5
    Hi again

    Tnx for all the math.
    I have all the constants except for the s, do you know what is the s for tap water?
    didn't check yet the math.

  7. Apr 15, 2012 #6


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    4200 Joules per litre per deg C
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