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I Difference between using temperature in Celsius and Kelvin

  1. Aug 27, 2016 #1
    For one my homework questions I was asked to find the specific heat capacitance of a certain material. The specific heat capacitance has a temperature dependence given by:
    c = 0.20 + 0.14T + 0.023T2
    the units for T is temperature in Celsius and c is measured in cal/gK.
    2.0 g of the material was to be raised from 5.0 Celsius to 15.0 Celsius.

    I did not have any difficulty solving the problem, as it was a fairly easy integral. I am just very uncomfortable that my first instinct was to convert the temperature to Kelvin. Why is this incorrect? If I stay in Celsius, I get an answer of 82 cal, but if I convert temperature to Kelvin (which I want to do because of the units of the specific heat) I get an answer of 38 kcal...which is very wrong.

    I always pay attention to the units I am working in to make sure everything works out correctly, but in this case it is not reasonable to do so and this is very unsettling to me. Could someone please explain why this is the case?

    Thank you in advance for any answers.
     
  2. jcsd
  3. Aug 27, 2016 #2

    fresh_42

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    It looks like the coefficients in the formula also carry (different) units, because otherwise ##1,T,T^2## couldn't be added.
    Changing to Kelvin requires them to be adjusted, too.
     
  4. Aug 27, 2016 #3

    Svein

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    1 degree difference Celsius equals 1 degree difference Kelvin.
     
  5. Aug 27, 2016 #4

    robphy

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    the symbol T likely means the change in temperature (final minus initial)... as opposed to the [absolute] temperature [in K].
    Then, as @Svein says, the change in Celsius is equal to the change in Kelvin.
     
  6. Aug 27, 2016 #5

    SteamKing

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    The equation for c appears to be derived from a curve fit to specific heat capacity data for this substance. Since K = C + 273 approximately, if you substitute Kelvin for Celsius temperature in this equation, you are going to get much higher values of the specific heat c. Some equations come with limitations on their use which arise out of how they were originally developed. You have to observe these limitations, otherwise the equation will evaluate to nonsense, as you have found.
     
  7. Aug 27, 2016 #6
    Thank you to everyone for your answers and explanations. I think it makes sense now as to why the conversion to Kelvin is more trouble than it is worth. I think I was just misinterpreting the symbol T, neglecting the fact that I may need to adjust the coefficients of T if I were to convert it.
     
  8. Aug 27, 2016 #7

    fresh_42

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    At least the coefficient at the ##T^2## term which isn't linear in ##T##, whereas ##C \rightarrow K## is. If ##T## denotes a difference in temperature, then this term will have to change non-trivially.
     
  9. Aug 28, 2016 #8
    Replace T in your equation for c by (T-273), collect terms, and then integrate between 278 and 288. I guarantee you will get 82 calories.
     
    Last edited: Aug 29, 2016
  10. Aug 28, 2016 #9

    sophiecentaur

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    You are worrying needlessly about dimensional analysis here. c is a dimensionless coefficient which varies, according to the value of T and that quadratic formula gives a reasonable approximation to the actual relationship between c and T. This technique is used in many instances to approximate a non-linear relationship between two quantities.
    c is used in the formula
    Q = cm ΔT so its units of c would be be Joules/kg/°C only if c was independent of temperature. There is an inherent error in any actual value of c that's used.
    The formula for c should have the temperature scale specified and the OP contains that information. All other temperature values are, in fact ΔT and that may or may not involve integration (depending on the error in c over the temperature range and wether it is significantly non linear over the range.)
     
  11. Aug 28, 2016 #10
    This is not quite true.
    There are well known cases where c depends on the temperature and the unit is the same (J/kg/C)
    See for example the specific heat of solids at low temperatures.
    The specific heat of metals at low temperature for example in many cases can be approximated by a polynomial containing cubic (due to phonons) and linear (due to free electrons) terms in temperature.
    Of course in these cases you cannot calculate the heat by the formula above but you need an integral.
    However this has no relevance for the units.
    Here is an example for the T3 behaviour. See units on the vertical axis. They use molar instead of specific (per kg) heat.
    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/imgheat/silisph.gif
     
  12. Aug 29, 2016 #11

    sophiecentaur

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    I thought I may have been on dodgy ground here but I suppose dimensional analysis doesn't have to apply with polynomial approximations to a non linear law.
    If there is a relationship:
    y = a exp(sin(log(x))), there must be a unit for the coefficient a but it certainly won't be yunits/xunits so where can you take it in dimensional analysis?
     
  13. Aug 29, 2016 #12
    I don't see where is the problem with dimensions in these cases. Each coefficient can have the proper dimensions in order to make the equation consistent.
    For your example, the unit for a is the same as the unit for y. And x is unit less.
    So you could say that
    aunit=yunit/xunit as the terms in the denominator is 1 but is not really relevant. It is clear that a and y must have the same unit, like in
    x=Asin(w*t) ; v=wA cos(w*t)
    for a harmonic oscillator.
    x and A have same dimension, length.
    v has same dimension as w*A. It does not matter what is inside the sin.

    If you expand the nonlinear part into a power series in x, all coefficients in the expansion will be dimensionless as x is dimensionless.
    But for a physical problem, x may be the ratio of two quantities with units, like x=T/To. And one may want to have the expansion in terms of T rather than x. This is also OK but now the coefficients will have different dimensions, for each term. Unfortunately in many technical sources the coefficients are given without units and this can create the impression that they are unitless, like in this formula for speed of sound versus temperature, for example:
    http://resource.npl.co.uk/acoustics/techguides/soundpurewater/content.html
     
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