Question #1c6ef

1 Answer
Nov 6, 2016

Here's what I got.


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# minutes, i.e. #60 xx "60 s"#, you will 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 #"504 kJ"# of heat will be

#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 #"1.0 g mL"^(-1)#. This means that the volume of water that can be evaporated will be

#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.