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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:

    c={\partial C \over \partial m}

    In absence of phase transition you have

    c=E_ m={C \over m} = {C \over {\rho V}}


    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:


    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:



  4. Mar 11, 2009 #3


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