# Question #4fbf8

##### 2 Answers

#### Answer:

Sorry...I cancelled my answer by mistake!

Ok, I found

#### Explanation:

Water uses the heat (=energy) to change its phase from liquid to gas maintaining the temperature constant at

So basically you give heat

Rearranging the above relationship you get:

#### Answer:

#### Explanation:

So, you know that you're dealing with a mass of

In order to be able to calculate how much steam is produced by adding *enthalpy of vaporization*,

To make the calculations a little more interesting, I'll use the value of *kilojoules per mole*

#DeltaH_"vap" = "40.657 kJ/mol"#

So, use the equation

#q = n_"water" * DeltaH_"vap"#

to figure out how much water can be converted to steam by the addition of

Use the conversion factor

#"1 kcal" = "4.184 kJ"#

to get

#5.40color(red)(cancel(color(black)("kcal"))) * "4.184 kJ"/(1color(red)(cancel(color(black)("kcal")))) = "22.6 kJ"#

This means that you have

#"22.6 kJ" = n_"water" * DeltaH_"vap"#

#m_"water" = (22.6color(red)(cancel(color(black)("kJ"))))/(40.657color(red)(cancel(color(black)("kJ")))/"mol") = "0.556 moles water"#

Use water's molar mass to figure out how many grams would contain this many moles

#0.556color(red)(cancel(color(black)("moles"))) * "18.02 g"/(1color(red)(cancel(color(black)("mole")))) = "10.01 g"#

Therefore, out of your initial

If you want, you can express that in *kilograms* to get

#10.0color(red)(cancel(color(black)("g"))) * "1 kg"/(1000color(red)(cancel(color(black)("g")))) = color(green)(1.00 * 10^(-2)"kg")#

**SIDE NOTE** *So, how much heat would be needed to convert all the water to steam?*

#1.00color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole"/(18.02color(red)(cancel(color(black)("g")))) = "55.5 moles"#

#q = 55.5color(red)(cancel(color(black)("moles"))) * 40.657"kJ"/color(red)(cancel(color(black)("mole"))) = "2256 kJ"#

*Finally, this is equivalent to*

#2256color(red)(cancel(color(black)("kJ"))) * "1 kcal"/(4.184color(red)(cancel(color(black)("kJ")))) = "539 kcal"#