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Heat energy required

  1. Mar 11, 2009 #1
    Hi, wonder somone could help me, Im not physics expert. If a lube oil has a specific heat of 0.444Btu/Lb.F, how do I calculate the heat energy (kw) required to raise its temperature from 10C to 40C. Amount of liquid is 35L. Will I be able to calculate the time it takes to raise the temperature as well? someone told me its as easy as using the equation Q=CMΔT but another said I have to use Q=cpdT, Im confused!
     
  2. jcsd
  3. Mar 11, 2009 #2
    First i think you should specify wich phisycal quantities hide behind those letters :-),
    in any case The specific heat capacity of a material is:

    [tex]
    c={\partial C \over \partial m}
    [/tex]

    In absence of phase transition you have

    [tex]
    c=E_ m={C \over m} = {C \over {\rho V}}
    [/tex]

    where:

    C is the heat capacity of a body made of the material in question,
    m is the mass of the body,
    V is the volume of the body, and

    [tex]\rho = \frac{m}{V} [/tex] is the density of the material.

    I bet this is the relation in your's formula:

    [tex]
    c_p=CM
    [/tex]

    where C is specific heat capacity at const pressure M the mass of the system and c_p is the body heat capacity at constant pressure.


    I hope this answer can help you, in any case you can check this page:


    http://en.wikipedia.org/wiki/Specific_heat_capacity

    bye

    marco
     
  4. Mar 11, 2009 #3

    russ_watters

    User Avatar

    Staff: Mentor

    Use this one:
     
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