"Calculate the enthalpy of vaporization of water given that, when 5.00 grams of steam (gaseous H2O) at 100°C was introduced into a beaker containing 500 grams of water at 20.0°C, the temperature rose to 26.2°C."

q_H2O(l) + q_H2O(g)= E

nΔH_vap + m_gasCΔT + m_H2O(l)CΔT = E

(500g * 4.18J/g°C * (26.2°C-20°C)) + [(500g/18g/mol) * (40.7kJ/mol * 1000J)] + (5g * 2.0J/g°C * (26.2°C-100°C)) = E ???

Not sure if this is the right way, but help is much appreciated.

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  1. The first term you have written is correct; the second and third are not.
    The first term correctly identifies the energy needed to raise T of 500 g H2O from 20 to 26.2. Your third term should be the steam at 100 C condenses to liquid water at 100 C and helps heat the water to 26.2. That is
    [5 x 4.184 x (26.2-100)]. Note that the only error you made here is you used the specific heat of steam and not water.

    Your second term is for the delta H vap. You can't use 40.7 kJ/mol. That's what you're trying to find. It is (5 x heat vap)

    Add these three terms and the sum = 0. I get about 2280 J/g

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