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

1. Jun 6, 2017

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
T(°F)=1.8⋅T(°C)+32

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.

Thank you

2. Jun 6, 2017

### TSny

Hello.

Check the part where you're converting the temperatures.

3. Jun 6, 2017

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
4. Jun 6, 2017

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. Jun 6, 2017

### TSny

This is correct. But this is not what you used in your conversion in the first post.

6. Jun 6, 2017

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. Jun 6, 2017

### TSny

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.

8. Jun 6, 2017

I'm not sure I follow. Won't the 9/5 stay with the T(C) ?

9. Jun 6, 2017

### TSny

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.

10. Jun 6, 2017