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

  • Thread starter AnotherParadox
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In summary: Thank you for your help!In summary, using dimensional analysis and the given conversion factors of 1 Btu = 1.055056 kJ and 1 kg = 2.2046226 lbm, it can be shown that 1 kJ/(kg⋅°C) is equivalent to 0.238846 Btu/(lbm⋅°F) by multiplying the units 1 kJ/(kg⋅°C) by 1 Btu / 1.055056 kJ and 1 kg / 2.2046226 lbm. However, when converting the temperature units, it is important to note that a change of 1 Co on the Celsius scale is equivalent
  • #1
AnotherParadox
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3

Homework Statement


[/B]
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
T(°F)=1.8⋅T(°C)+32

Homework Equations


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

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
 
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  • #2
Hello.

Check the part where you're converting the temperatures.
 
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  • #3
TSny said:
Hello.

Check the part where you're converting the temperatures.

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:
  • #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
 
  • #5
AnotherParadox said:
1.8 Δ°F = 1 Δ°C
This is correct. But this is not what you used in your conversion in the first post.
 
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  • #6
TSny said:
This is correct. But this is not what you used in your conversion in the first post.
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 .
 
  • #7
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.
 
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  • #8
TSny said:
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.
I'm not sure I follow. Won't the 9/5 stay with the T(C) ?
 
  • #9
AnotherParadox said:
I'm not sure I follow. Won't the 9/5 stay with the T(C) ?
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.
 
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  • #10
The suppressed units is what got me. Makes perfect sense now
 

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

1. How do I convert kJ/(kg*degC) to Btu/(lbm*degF)?

To convert kJ/(kg*degC) to Btu/(lbm*degF), you can use the following formula: Btu/(lbm*degF) = (kJ/(kg*degC) * 0.4299. This will give you the equivalent value in Btu/(lbm*degF).

2. What is the conversion factor for kJ/(kg*degC) to Btu/(lbm*degF)?

The conversion factor for kJ/(kg*degC) to Btu/(lbm*degF) is 0.4299. This factor is used to convert the units from the metric system (kJ/(kg*degC)) to the imperial system (Btu/(lbm*degF)).

3. Can I use an online converter for this conversion?

Yes, there are many online converters available that can easily convert kJ/(kg*degC) to Btu/(lbm*degF). Simply input the value in kJ/(kg*degC) and the converter will give you the equivalent value in Btu/(lbm*degF).

4. How is kJ/(kg*degC) different from Btu/(lbm*degF)?

kJ/(kg*degC) and Btu/(lbm*degF) are two different units used to measure specific heat capacity. kJ/(kg*degC) is used in the metric system, while Btu/(lbm*degF) is used in the imperial system. They both measure the amount of heat required to raise the temperature of a substance by 1 degree Celsius or 1 degree Fahrenheit respectively.

5. Why is it important to know how to convert kJ/(kg*degC) to Btu/(lbm*degF)?

Knowing how to convert units from one system to another is important for scientific research and communication. It allows scientists to compare and understand data that may use different units of measurement. Additionally, some experiments may use different units, so being able to convert between them is necessary for accurate results.

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