How to sort ten coin for twenty single weightings? We can sorting 10 coin for 20 weighting, don't it? We can know which of the two coins is heavier for one weighting. Any two coin have different weight.

I interpret the question to be whether we can determine the order of the weights of $10$ coins using $20$ weighings.

The answer is no, since there are $10!=3628800$ different orders and only $2^{20}=1048576$ different results of the $20$ weighings. You need at least $22$ weighings (and probably quite a bit more).