Re: help please with advanced conversions

Re: help please with advanced conversions   (04/23/2007, 10:55 AM)

1.158 J/(g.K) to J/(g.°C).

°C = K - 273,15

But since we're talking about the amount of heat needed to raise the temperature of one gram of a substance by one Kelvin, it is on a "per Kelvin" basis, and thus the starting point of the scale is irrelevant.

1.158 J/g-K = 1.158 J/g-°C.