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Thermodynamics/Cooling Problem ^_^

  1. Oct 2, 2006 #1
    A homeowner finds that, if she turns down her thermostat by 3[degrees C] she can reduce the rate at which heat is lost to outside, by 10%. If the outside temperature is -5[degrees C], what are the Thermostat readings before and after the adjustment.

    I'd just like verification that i haven't screwed up this problem totally...
    [because i'm just sorta going along blindly, moving on when I touch something...]

    My teacher gave me this equation:

    d Q/ d t{ime} = (T{emp hot} - T{emp cold}) * A{surface area} * K(conductivity constant) / L{ength of conducted material}

    dQ/dt = (T(h) - T(c)) * A * K / L


    While i know this equation:
    Q = mc/\T

    However, i chose to use sole my teacher-given equation...
    dQ/dt = (T(h) - T(c)) * A * K / L

    and instead equated it to
    /\Q//\t = (T(h) - T(c)) * A * K / L
    (i used change in heat/time, i'm not so good with calculus :tongue: )

    /\Q//\t = (T(h) - T(c)) * A * K / L
    I decided that the orange terms would be kept constant.

    Constant = (T(h) - T(c)) / (/\Q)

    then i chose two thermostat readings to be equal to that constant

    (Before thermostat change) Constant = (To - (-5) / (Q)

    (After thermostat change) Constant = ((To - 3) - (-5) / (.9Q)

    I then set the two equations equal to each other, via the constants, cancelled out the Q on both sides, Solved for To,
    and got

    75[degrees C] for the beginning Temperature...

    Any suggestions?
     
  2. jcsd
  3. Oct 2, 2006 #2
    *bump for attention*
     
  4. Oct 2, 2006 #3
    Okay, since your teacher gave you the function, then I assume it is alright if I derive it again because it is out of your scope. The formula you have deals with conductivity and I don't know any conductive houses.

    We know
    [tex]\frac{dT}{dt}=kT_e-kT[/tex] T_e = environment and k>0
    Which can be written as
    [tex]\frac{dT}{dt}+kT=kT_e[/tex]
    so, we can solve with an integrating factor
    [tex]\mu (t) = e^{\Int k dt} = e^{kt+c}[/tex]
    Then we can multiply the original differential by the integrating factor, where the constant on the exponential will factor out
    [tex] \mu (t)[ \frac{dT}{dt} + kT = kT_e][/tex]
    Then simplify the left hand side, and we can integrate both sides so that
    [tex] \int \frac{d}{dt}(\mu (t) T(t)) = \int kT_e* \mu (t) dt[/tex]
    Which will give, after plugging the value for u(t) back in and solving
    [tex] e^{kt}T(t) = T_e e^{kt} + C[/tex]
    Finally, multiply both sides by e^-kt to get
    [tex]T(t) = T_e + C e^{-kt}[/tex]

    That is your formula, which has two constants that can be solved for separately with two initial values. So T(t) is the temperature dependent on time, which relates to one of your givens (I'll let you try to figure which one out). Though there is one thing I am a little confused on. It could be that you are supposed to leave the answer in terms of variables or something, but without an initial temperature in the house I don't know how you are supposed to figure out the constant. You didn't leave anything out did you?
     
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