# Question #1c6ef

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The first thing to do here is to figure out how many *joules* of energy would be produced by your lightbulb in *one hour*.

As you can see, the conversion factor to use here is

#"1 W" = "1 J"/"1 s"#

This is basically a reminder of the fact that the *watt*, a unit of **power**, is defined as an energy of *one joule* delivered in *one second*.

Since one hour is known to have

#60 xx 60 color(red)(cancel(color(black)("s"))) * "140 J"/(1color(red)(cancel(color(black)("s")))) = "504000 J" = "504 kJ"#

Your next step here will be to use the **enthalpy of vaporization** of water to figure out how many *grams* of water would be evaporated by that much heat.

#DeltaH_"vap" = "44.66 kJ mol"^(-1)#

Convert this to *kilojoules per gram* by using water's **molar mass**

#44.66 "kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole H"_2"O"))))/("18.015 g") = "2.479 kJ g"^(-1)#

This means that the mass of water that can be evaporated by

#504color(red)(cancel(color(black)("kJ"))) * ("1 g H"_2"O")/(2.497color(red)(cancel(color(black)("kJ")))) = "201.8 g"#

Now, *assuming* that the sweat is *pure water*, you can approximate its density to be equal to

#201.8 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.0color(red)(cancel(color(black)("g")))) = color(green)(bar(ul(|color(white)(a/a)color(black)(2.0 * 10^2 "mL")color(white)(a/a)|)))#

The answer is rounded to two **sig figs**.