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Suppose that 1.01g of rubbing alcohol(C3H8O) evaporates from a 56.0g 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


    1. Look up the heat of vaporization of isopropyl alcohol. The figure I found is 44.0 kJ/mole. Divide that by the molar mass to get the kJ/g. Multiply by 1.01g to get the heat absorbed from the aluminum block.

    2. The applicable formula for the aluminum is:
    heat = (specific heat of Al)(grams Al)(deltaT). Look up the specific heat of Al. Solve for deltaT, the temp. change. Subtract that from 25.0ºC.

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