He added in a note that the entry should have been titled -262.15 ºC, but as we can fully understand, he was typing very quickly, knowing that victory might escape him. Now can anybody tell me what happens at -262.15 ºC?
By
Ned Gulley
Ned is part of the MATLAB Central team.
0 K = -273.15 °C
11 K = -262.15 °C
i guess :)
Ah! Very good! I was trying to work out the liquefaction temperature of various gases, and it was getting me nowhere. Thanks for the insight.