On 8 January 2014 10:10, Andreas F. Geissbuehler <[email protected]> wrote: >> ...you would need 70,000kcal/degree to melt... > > If you can't wait until July, 35,000 Cal per degree (C|K) > should do - kcal / Cal / Kal = kilo-calorie > 1 Cal = 1kg water / degree (C|K) > > Hopefully I made my point in favor of SI !
Except that the Calorie (kilo or otherwise) is not an SI unit - not even a derived one. That would be the Joule, AKA Newton metre (Nm - not to be confused with nautical mile) or Watt second.. At the wholesale level, natural gas and electricity are quoted in GJ or TJ. And some food labels show both kCalories and kJ, though these are specific to human food consumption and not easily related to thermal value. (A kg of grass has almost 0 Calories for a human, but lots for a cow.) Are we sufficiently Friday-digressed yet...? Tony H.
