Suppose that 1.15g of rubbing alcohol (C3H8O) evaporates from a 67.0g aluminum block. If the aluminum block is?

If the aluminum block is initially at 25, degrees celsius 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 degrees celsius. Heat of vaporization of the alcohol at 25 degrees celsius is 45.4 kJ/mol?

Help?!

1 Answer

  • (1.15 g C3H8O) / (60.09535 g C3H8O/mol) x (45.4 kJ/mol) =

    868.79 J gained by the alcohol and lost by the Al

    To continue this problem you need to know the specific heat of Al.

    The source below says 0.900 J/g·°C.

    (868.79 J) / (0.900 J/g·°C) / (67.0 g) = 14.4 °C change

    25 °C - 14.4 °C = 11 °C

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