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Suppose that 1.08 g of rubbing alcohol (C3H8O) evaporates from a 65.0 g aluminum block.?
Suppose that 1.08 g of rubbing alcohol (C3H8O) evaporates from a 65.0 g aluminum block.
If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. Heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol
I keep getting this wrong, I don t know why.
1 Answer
- Roger the MoleLv 74 years agoFavorite Answer
(1.08 g C3H8O) / (60.0950 g C3H8O/mol) x (45.4 kJ/mol) = 0.81591 kJ =
815.91 J gained by the alcohol (and lost by the aluminum)
(815.91 J) / (897 J/kg·°C) / (0.0650 kg Al) = 14.0°C change
25°C - 14.0°C = 11°C
