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Math Help - I am trying to find out two unknown values from one equation

  1. #1
    SOS
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    I am trying to find out two unknown values from one equation

    Does anyone know how to find two unknown values in the one equation it is for a year twelve maths b assignment. I am trying to find out what c and k are in this equation (by the way c and k are constants):
    T=TS + ce^(-kt)
    Where TS= Temperature of surroundings which equals 23 degrees
    T= Temperature of the body at a time (t)
    t= Time
    c & k= constants
    Thank you for your help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by SOS View Post
    Does anyone know how to find two unknown values in the one equation it is for a year twelve maths b assignment. I am trying to find out what c and k are in this equation (by the way c and k are constants):
    T=TS + ce^(-kt)
    Where TS= Temperature of surroundings which equals 23 degrees
    T= Temperature of the body at a time (t)
    t= Time
    c & k= constants
    Thank you for your help.
    You need data, say some times t_1, .. t_n, and the corresponding tempratures T(t_1), .. T(t_n).

    RonL
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  3. #3
    SOS
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    I have Values for T and t but I don't know how to find the 2 unknowns in the 1 eqn

    thanks ronL, in class we got the time and temperature which is:
    t= 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
    T= 93 75 62 51 45 41 37 35 33 31 30 29 28 27.8 27
    But even if I substitute those numbers in I still have two unknowns left, and I don't know what to do after I substitute T and t in.
    Thank you for your help
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by SOS View Post
    Does anyone know how to find two unknown values in the one equation it is for a year twelve maths b assignment. I am trying to find out what c and k are in this equation (by the way c and k are constants):
    T=TS + ce^(-kt)
    Where TS= Temperature of surroundings which equals 23 degrees
    T= Temperature of the body at a time (t)
    t= Time
    c & k= constants
    Thank you for your help.
    Check your problem again...these types of problems usually come with initial conditions such as

    when t= y=
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  5. #5
    SOS
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    The question in full

    ok umm.. this is the question in full:
    "Newton's Law of Cooling states that the rate of change of temperature of a body is proportional to the difference in temperature between the body and the temperature of its surrounds."
    (I don't actually understand the above sentence is that what you mean by t= y=)
    "The formula fo Newton's law of cooling is:
    T=TS + ce^(-kt)
    Where TS= Temperature of surroundings which equals 23 degrees

    T= Temperature of the body at a time (t)
    t= Time
    c & k= constants"
    I then have made a graph of time (t) verse temperature (T) when t=x and T=y
    It then says to
    "Determine how well Newton's Law of Cooling models your experimental data. Your solution should include the use of a spreadsheet and a comparative graph."
    So I thought that it meant that I had to come up with values for the constants so that I could model the formula against the data I obtained in a graph.
    Thanks for your help.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by SOS View Post
    thanks ronL, in class we got the time and temperature which is:
    t= 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
    T= 93 75 62 51 45 41 37 35 33 31 30 29 28 27.8 27
    But even if I substitute those numbers in I still have two unknowns left, and I don't know what to do after I substitute T and t in.
    Thank you for your help
    You wish to fit a curve of the form:

    T(t)=T_s+ce^{-kt}

    to this data.

    Rearrange:

    (T(t)-T_s)=c e^{-kt}

    Now take logs to get:

    \log(T(t)-T_s)=-kt+\log(c)

    so plot t against u(t)=\log(T(t)-T_s) , this should be a straight line and its slope will be -k, and the intercept \log(c).

    RonL
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