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Thermodynamics: Third World Bacteriological Test

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data
    For bacteriological testing of water supplies and in medical clinics, samples must routinely be incubated for 24 h at 37°C. A standard constant temperature bath with electric heating and thermostatic control is not suitable in developing nations without continuously operating electric power lines. Peace Corps volunteer and MIT engineer Amy Smith invented a low cost, low maintenance incubator to fill the need. The device consists of a foam-insulated box containing several packets of a waxy material that melts at 37°C, interspersed among tubes, dishes, or bottles containing the test samples and growth medium (food for bacteria). Outside the box, the waxy material is first melted by a stove or solar energy collector. Then it is put into the box to keep the test samples warm as it solidifies. The heat of fusion of the phase-change material is 205 kJ/kg. Model the insulation as a panel with surface area 0.490 m2, thickness 4.50 cm, and conductivity 0.0120 W/m·°C. Assume the exterior temperature is 23°C for 12.0 h and 16°C for 12.0 h. (a) What mass of the waxy material is required to conduct the bacteriological test? (b) Why are the mass of the samples and the mass of the insulation not factored into the calculations?

    2. Relevant equations
    [tex] P = \dfrac{kA \left( T_h - T_c \right)}{L} [/tex]
    [tex] Q = L_f \Delta m [/tex]



    3. The attempt at a solution
    So I figure that in the 24 hours, the amount of heat (assuming there is enough) that will transfer through the "panel" (escaping from inside the container) will be equal to the rate of heat transfer x time:
    [tex] Q = \left( \dfrac{kA \left( 37-23^{\circ}C \right) }{L} + \dfrac{kA \left( 37-16 ^{\circ} \right) }{L} \right) \cdot \left( 3600 \dfrac{s}{h} \cdot 12 \right) = L_f \Delta m[/tex]
    And solving for m:
    [tex] m = \left( 3600 \dfrac{s}{h} \cdot 12 \right) kA \dfrac{74-23-16^{\circ}C}{L \cdot L_f}[/tex]
    Where:
    L = .0450 m
    Lf = 205 kJ/kg = 2.05e5 J/kg
    k = .0120 W/m2
    A= .490 m2

    I get m = 0.96 kg

    There is no solution to this problem in my book and my professor (bless his heart?) posted solutions to an old edition. So... does this look right? Also, I can't figure out a good answer to part (b) other than, if the samples are in thermal equilibrium at the beginning of the 24 hours, as long as the rate of energy is steady state the temperature will not change and the bacteria will be sufficiently warmed.
     
    Last edited: Nov 25, 2012
  2. jcsd
  3. Nov 25, 2012 #2

    haruspex

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    I guess that should read LfΔm.
    There are two 12 hour periods, but you have a time factor of 24 hours for each rate.
    Other than that, looks fine.
    I can't find the question for part (b).
     
  4. Nov 25, 2012 #3
    Ah, I overlooked that. Thank you, and yes it was supposed to be L_f! Part (b) is why the mass of the sample and the mass of the insulation are not factored into the calculations.
     
  5. Nov 25, 2012 #4

    haruspex

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    What you answered (in particular, if the samples are already below temperature presumably they're spoilt), plus, the insulation will have relatively low specific heat.
     
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