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British Thermal Unit Definitions 1°F/°C

Note that the definitions of the BTU included references to 1°F/°C. This is erroneous. The actual correct mathematical multiplier is (1°F - 0°F) / (1°C - 0°C). This is equivalent to 1°R/K which is not equivalent to 1°F/°C. 1°R/K = 0.55555555555556; however, 1°F/°C = 0.93353192696618. Please comment if you you can prove this is not the case. —Preceding unsigned comment added by 70.219.195.46 (talk) 05:18, 4 March 2008 (UTC)

First of all, I fail to see how either of the two BTU "definitions" were actually definitions, so I removed them. As I understand it, the BTU is defined as the temperature required to raise one pound of water by one degree Fahrenheit. The different versions are due to different starting temperatures for the water.
If any evidence is ever provided that the definitions were correct, then we could consider whether 1 °F/ °C is equivalent to 1 °R/ K. Since all are being used to measure temperature intervals, they are equivalent. 1 °F / 1 °C = 9/5 = 1.8, not 0.93353192696618. --Gerry Ashton (talk) 16:19, 4 March 2008 (UTC)
I beg to differ with you. 1 °F = 255.92777777778 K and 1 °C = 274.15 K therefore 1 °F / 1 °C = 255.92777777778 K / 274.15 K = 0.93353192696618; however, 1 °R/ K = 0.55555555555556 K / 1 K = 0.55555555555556. Remember that °F and °C are relative scalers and must be converted to absolute to do any meaningful mathematics. Unless of course you want to do relative math ( In which case 1 °F = -17.222222222222 °C which means 1 °F / 1 °C = -17.222222222222 that's just silly )
You are correct in stating that °F and °C are suitable for denoting temperature intervals; however, referring to the Wikipedia entry for Celsius sub-section Temperatures and Intervals: "Because of this dual usage, one must not rely upon the unit name or its symbol to denote that a quantity is a temperature interval; it must be unambiguous through context or explicit statement that the quantity is an interval" --Richard Wolfgram 00:15, 5 March 2008 (UTC)
The British Thermal Unit is defined as the amount of energy required to raise the temperature from one temperature to a temperature one degree Fahrenheit higher. This is a temperature interval. Any BTU will be roughly the same amount of energy so long as the water is a liquid at both the starting and ending temperature, whether the increase is from 32 °F to 33 °F, or from 63 °F to 64 °F, or any temperatures in the liquid range, but for greater precision, the starting temperature and all other parameters for the operation must be specified. It should be unambiguous to anyone who follows the link to the full explaination of British Thermal Unit and reads (and understands) the definition that a temperature interval is being used.
Finally, notice that I deleted the "definitions" altogether, because the "definitions" that used °F and °C were not really definitions either. --Gerry Ashton (talk) 02:08, 5 March 2008 (UTC)
I'll give you that 1 °F/°C it can be deduced to mean an interval. I still consider it poor math. Consider the Gas Law Formulas P1 * V1 = (P2 * V2) * (T2/T1). Solving this you could easily end up with 1 °F/°C = 0.93353192696618. It would be poor math to not assume absolute temperature.
I thought the thermalchemical BTU was based on the thermalchemical calorie; and, I apologize if I pressed the issue beyond a reasonable relevant discussion.
Just for kicks consider 1°F + 1°C = ? then consider 1°C + 1°F = ? --Richard Wolfgram 23:00, 5 March 2008 (UTC)
For a while, there was a distinction between degF (an interval), and °F (a point). In practice, one now uses °F for both. So one can add 1°F to 1°C to get 2.8°F, since one must suppose both are intervals. Absolute interfals might be expressed in rankines and kelvins.
In any case, one should not do things like convert eg the 60°F in BTUKsub>60°F, since the imperial and metric scales define different ranges of temperature. For example, 68°F runs from 68°F to 69°F, while the equal 20°C runs from 67.1°F to 68.9°F, and the specific heats of water for 68°F is thus different to 20°C.
One should note that the various BTUs and calories are actually derived units, the relation being W = j.MΘ. It's the value of the joule constant that is defined, usually by the metric value. So a BTUIT is 1 lb × 1°F × jIT, where the last value is 4186.8 J/kg.K or 2326 J/kg.°F. Ice, as in (ton-ice/day), is defined as 144°F (in any sub-scale), so there is something like iceIT = 144 × 2326 J/kg or 80 K × 4186.8 J/kg KWendy.krieger (talk) 12:32, 4 August 2012 (UTC)

Regulo Gas Marks

I added this to the temperature scale, since it is thus described in the Gas-Mark page, and used without comment in various cookbooks. The definition is given in the remarks as degF = GM×25 + 300, and the formula is into kelvins. Wendy.krieger (talk) 12:35, 4 August 2012 (UTC)

I would like to post a link to an Excel spreadsheet providing user defined functions with the following features:

  • Free and open source
  • Conversion of units using editable tables based on the Wikipedia data
  • Unit aware evaluation of formulae entered as text
  • Open Source VBA functions that can be used as user defined functions on the spreadsheet or called from other VBA routines

As far as I know this is the only Excel spreadsheet available with these features.

The link is Units4Excel.

Any comments? How do I go about posting the link?

DougAJ4 (talk) 22:59, 28 August 2012 (UTC)

What evidence is there you are an expert in this field. If you read the WP:External links guideline, you will see that although external links don't have to meet all the criteria of reliable sources, they should have been created by knowledgeable people. There should be some evidence of that knowledge. Jc3s5h (talk) 00:51, 29 August 2012 (UTC)


What evidence would you like? Have a look at the spreadsheet and check that it works, and that it does what it says it does. You can look at the other spreadsheets available on the site for evidence of expertise in VBA. Also see:

for evidence of other Wikipedia links to spreadsheets on my site.

DougAJ4 (talk) 02:24, 29 August 2012 (UTC)

Evidence from an impartial source. I don't load VBA from strangers. Jc3s5h (talk) 02:28, 29 August 2012 (UTC)
If you had a look at the contents of the site and the comments it should be obvious that I know how to program VBA and that the applications I post are useful, non-malicious and non-commercial. Further evidence is:

DougAJ4 (talk) 02:56, 29 August 2012 (UTC)

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