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Help converting kJ/(kg*degC) to Btu/(lbm*degF)

  1. Jun 6, 2017 #1
    1. The problem statement, all variables and given/known data

    Show that 1 kJ/(kg⋅°C) = 0.238846 Btu/(lbm⋅°F) using dimensional analysis and the given conversion factors:

    1 Btu = 1.055056 kJ
    1 kg = 2.2046226 lbm

    2. Relevant equations

    1 Btu / 1.055056 kJ =1
    1 kg / 2.2046226 lbm = 1
    1.8⋅Δ°C / Δ°F=1

    3. The attempt at a solution

    (1 kJ/(kg°C)) ⋅(1 Btu / 1.055056 kJ) ⋅ (1 kg / 2.2046226 lbm) ⋅ (1.8⋅Δ°C / Δ°F) ≈ 0.842661 Btu/(lbm⋅°F)

    I have no idea how this is wrong but it is according any conversion table I've seen.

    Please help, I need to know what I'm doing wrong.

    Thank you
  2. jcsd
  3. Jun 6, 2017 #2


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    Check the part where you're converting the temperatures.
  4. Jun 6, 2017 #3
    Edit: I think I know what I'm doing wrong but still need help, ready below this post.

    How is it wrong ? I've been staring at it for hours. The change is 1.8 or 9/5 ..

    100 C = 212 F
    0 C = 32 F
    The slope is 9/5 or 1.8 with a y intercept of 32. In the analysis wouldn't this be multiplying by a factor of 1.8 since I'm going from inverse C to Inverse F.. I may be confusing myself with the methodology here idk
    Last edited: Jun 6, 2017
  5. Jun 6, 2017 #4
    Sorry for double post

    So I think I understand what's wrong

    The "change in F" and "change in C" is what I'm comparing

    It makes sense that a change of 1.8 F only results in a change in 1 C so I can change my conversion factor... but how do I show this logic rigorously or in math terms? In other words.. how do I show more work stating from the T(F) = 9/5 T(C) + 32 results in 1.8 Δ°F = 1 Δ°C
  6. Jun 6, 2017 #5


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    This is correct. But this is not what you used in your conversion in the first post.
  7. Jun 6, 2017 #6
    Right. Once I replace the conversion factor the answer is correct.

    However I'm still unsure how to take the information
    T(F) = 9/5 T(C) + 32
    and conclude
    1.8 Δ°F = 1 Δ°C

    in a way other than just mental reasoning ... how do I apply math notation to this to show something rigorous? Perhaps I'm concerned out of illusion .
  8. Jun 6, 2017 #7


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    Consider an arbitrary initial temperature and an arbitrary final temperature. So,

    Tfinal(F) = 9/5 Tfinal(C) + 32

    Tinitial(F) = 9/5 Tinitial(C) + 32

    Subtract the two equations.
  9. Jun 6, 2017 #8
    I'm not sure I follow. Won't the 9/5 stay with the T(C) ?
  10. Jun 6, 2017 #9


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    Yes. So you will get ΔT(F) = 9/5 ΔT(C). Now interpret this.

    Suppose you consider the case where the temperature change is 1 Co. The interpretation of the equation is that it tells us that this temperature change of 1 Co will be a change on the Fahrenheit scale of 9/5 Fahrenheit degrees.

    The confusion might be coming from the fact that the 9/5 has hidden units. The 9/5 has units of Fo/Co. So, when we consider a temperature change of 1 Co, we have

    ΔT(F) = (9/5 Fo/Co) ⋅ (1Co) = 9/5 Fo.

    So, 1 Co is equivalent to 9/5 Fo.
  11. Jun 6, 2017 #10
    The suppressed units is what got me. Makes perfect sense now
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