# How much heat input in #"kJ"# is required to produce #"450 g"# of water vapor from liquid water?

##### 1 Answer

You are referring to vaporization, a situation where heat flow **constant pressure** AND **constant temperature**.

So, this is asking for the **enthalpy of vaporization**,

When a substance vaporizes, its vapor pressure is constantly equal to the atmospheric pressure, and we have the following relationship between the **enthalpy** and the **heat flow**:

#color(green)(DeltaH = q_p)#

Naturally, if we divide by the

#(DeltaH_"vap")/(n_("H"_2"O")) = DeltabarH_"vap"^("H"_2"O") = q_p/(n_("H"_2"O"))#

The **molar enthalpy** *known* to be about

#\mathbf(q_p = n_("H"_2"O")DeltabarH_"vap"^("H"_2"O"))#

At this point, all we need to do is figure out the

#"mols H"_2"O" = "450" cancel("g H"_2"O") xx ("mol H"_2"O")/("18.015" cancel("g H"_2"O"))#

#= 24.979# #"mols"#

and now we can solve for the heat flow

#color(blue)(q_p) = "24.979" cancel("mols H"_2"O") xx "40.7 kJ"/cancel("mol H"_2"O")#

#=# #color(blue)("1016.65 kJ")#