Question #1c6ef
1 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.