Calculate the number of calories required to condense 45.0 g of steam at 100.°C.

Answer 1

To condense 45 g of steam at 100º requires nothing more than a phase change. We look up the ∆H of vaporization and use that value to calculate the energy liberated, not required.

∆Hvap =~540 cal/g

q = m x Hvap

q = (45.0 g)(540 cal/g)

q = 24,300 cal (3 sig. figs.)

Answer 2

Q= mc deltaT

You have m, you have c and I have no clue where it the other T

How can you condense something that will evaporate immediately?

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