Question

How much heat is needed to vaporize `"25.0 mL"` of water at its normal boiling point? `DeltaH_(vap) = "40.65 kJ/mol"`.

Answer 1

`"54.1 kJ"` into the water.

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Vaporizing at constant pressure (and of course, temperature) allows us to say:

`bb(q_(vap) = nDeltaH_(vap))`,

where `DeltaH_(vap)` is the change in enthalpy of vaporization, `n` is the mols of water, and `q_(vap)` is the heat flow through the water. For water, `DeltaH_(vap) = "40.65 kJ/mol"`.

Knowing that the density of water at `100^@ "C"` is `"0.9583665 g/mL"`:

`25.0 cancel"mL" xx "0.9583665 g"/cancel"mL" = 23.9_(591625)` `"g water"`

where the subscripts indicate digits past the last significant digit.

This amount of water corresponds to

`23.9_(591625) cancel"g water" xx "1 mol water"/(18.015 cancel"g water")`

`= ul(1.32_(9956)" mols of water")`.

So, the heat used to vaporize it was:

`color(blue)(q_(vap)) = 1.32_(9956)` `cancel"mols water" xx "40.65 kJ"/cancel"mol water"`

`=` `color(blue)("54.1 kJ")`, INTO the water,

to three sig figs.