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Math Help - writing equation for integration regarding rate of change

  1. #1
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    Smile writing equation for integration regarding rate of change

    Hi

    An object with temperature 26 degrees Celsius is placed in water with constant temperature of 90 degrees Celsius. If the temperature of the object rises to 70 degrees Celsius in five minutes, what will be the temperature after 10 minutes?

    I thought of using Newtons law of cooling d(temp)/d(t) = -k(constant of proportionality) (temp(object)-temp(surrounding medium)) but have read on the internet that it only describes transfer of heat from object to water not the other way around.

    Any ideas would be appreciated
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  2. #2
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    Quote Originally Posted by Poppy View Post
    Hi

    An object with temperature 26 degrees Celsius is placed in water with constant temperature of 90 degrees Celsius. If the temperature of the object rises to 70 degrees Celsius in five minutes, what will be the temperature after 10 minutes?

    I thought of using Newtons law of cooling d(temp)/d(t) = -k(constant of proportionality) (temp(object)-temp(surrounding medium)) but have read on the internet that it only describes transfer of heat from object to water not the other way around.

    Any ideas would be appreciated
    I suggest using the model \frac{dT}{dt} = k (90 - T) where T(5) = 70.
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  3. #3
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    formular

    I also posted this on another website and the reply was the same equation except it was dT/dt=k(T-90) not dT/dt=k(90-T). Could you please explain why yours is correct, I am a little confused haha

    Thanks
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  4. #4
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    Quote Originally Posted by Poppy View Post
    I also posted this on another website and the reply was the same equation except it was dT/dt=k(T-90) not dT/dt=k(90-T). Could you please explain why yours is correct, I am a little confused haha

    Thanks
    k represents two different things in the two DE's. In mine k > 0 (why?), in the other one k < 0 (why?).
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    k represents two different things in the two DE's. In mine k > 0 (why?), in the other one k < 0 (why?).
    Is it because the object gains energy as opposed to losing it (as in the original Newtons equation)? `

    Anyway, my integration does not seem to be right either

    dT/dt =k(90-T)
    dT =k(90-T) dt
    1/(90-T) dT=k dt

    -log(90-T)=kt+C
    log(90-T)=-kt0+C
    90-T=e(-kt)+C
    T=-e(-kt)+90+C

    70-90=-e(-k(5))+C
    log(20)/-5=k+C

    29-90=-e(-k(0))+C
    64=-1+C
    65=C

    (log(20)/-5)-65=k
    k=-65.59914645

    T=-e(65.59914645(10))+65
    T=-3.0856546...*10^28

    This cannot be right...where am i going wrong?
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  6. #6
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    my e ing was wrong

    -log(90-T)=kt+C
    log(90-T)=-kt0+C
    90-T=e(-kt)+C
    T=-e(-kt)+90+C

    becomes

    -log(90-T)=kt+C
    log(90-T)=-kt-C
    90-T=e(-kt-C)
    T=-e(-kt)*e(-C)+90 // e(-C)=K
    T=-Ke(-kt)+90
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  7. #7
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    Quote Originally Posted by Poppy View Post
    Is it because the object gains energy as opposed to losing it (as in the original Newtons equation)? `

    Anyway, my integration does not seem to be right either

    dT/dt =k(90-T)
    dT =k(90-T) dt
    1/(90-T) dT=k dt

    -log(90-T)=kt+C
    log(90-T)=-kt0+C
    90-T=e(-kt)+C Mr F says: You don't seem to realise that C is also meant to be in the exponent .... (Errrmmm, you do realise that e is raised to an exponent, don't you ....? Because what you have written suggests multiplication). Everything that follows will be wrong. Also, and not that it has consequences here, modulus signs should have been used when integrating to get the log: it should have been log |...| not log(....)

    T=-e(-kt)+90+C

    70-90=-e(-k(5))+C
    log(20)/-5=k+C

    29-90=-e(-k(0))+C
    64=-1+C
    65=C

    (log(20)/-5)-65=k
    k=-65.59914645

    T=-e(65.59914645(10))+65
    T=-3.0856546...*10^28

    This cannot be right...where am i going wrong?
    I'd say a lot of your trouble is handling the basic algebra of exponentials. You might be wise to review that material.
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  8. #8
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    about the exp

    sorry, my fault those were typos- well i typed them intentionally that way but thinking they ment other things(that is the way they are typed into matlab) haha
    Thanks for all your other help
    i managed to do the problem
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  9. #9
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    Quote Originally Posted by Poppy View Post
    Hi

    An object with temperature 26 degrees Celsius is placed in water with constant temperature of 90 degrees Celsius. If the temperature of the object rises to 70 degrees Celsius in five minutes, what will be the temperature after 10 minutes?

    I thought of using Newtons law of cooling d(temp)/d(t) = -k(constant of proportionality) (temp(object)-temp(surrounding medium)) but have read on the internet that it only describes transfer of heat from object to water not the other way around.

    Any ideas would be appreciated
    Heat always moves from the hotter body to the cooler. Whether heat moves from the object to the water or "the other way around" depends entirely upon whether the object or the water is hotter. You can use "Newton's law of cooling" in either case.
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