Question

What will be the final temperature of the water in an insulated container as the result of passing 5.00 g of steam [H2O(g)] at 100.0°C into 195.0 g of water at 20.0°C? (?H°vap = 40.6 kJ/mol H2O)


The best way to do these problems is to make the heat gained + heat lost = 0.
q1 = heat gained by water = mcdeltaT
massH2O x specific heat water x (Tf-Ti). Tf is the final T and Ti is the initial T.
q1 = 195.0 g x 4.184 J/g*C x (Tf - 20.0)
q2 = heat lost by steam in converting steam (g) @ 100.0 C to liquid @ 100.0 C = mass x delta H.
q2 = 5.00g x 40,600 Joules = ??
q3 = heat lost with 5.00 g (from the steam conversion)water moving from 100.0 C to final T.
q3 = 5.00g x 4.184 J/g*C x (Tf - Ti).
Ti is 100.0.
q1 + q2 + q3 = 0.
Plug in for q1, q2, and q3 and solve for Tf.

Answer 1

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