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heat requried to vaporize water

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How many grams of propane must be burned to supply the heat required to vaporize 0.680 L of water at 298 K?
C3H8(g) + 5 O2(g)-----> 3 CO2(g) + 4 H2O(l) Change of Heat = -2.22 x 10^3 kJ
H2O(l) ------> H2O(g) Change of Heat of Vapn. = 44.0 kJ

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Solution Summary

It finds the mass of propane that must be burned to supply the heat required to vaporize water at 298 K.

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(1) determine the amount of heat required to vaporize 0.680 L of water at 298K. That is obtained from the heat vaporizaton of water given in the problem:
H2O(l) ------> H2O(g) Change of Heat of Vapn. = 44.0 kJ
Remember that this is 44.0 kJ ...

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