Question #b91ab

1 Answer
Apr 10, 2017

Answer:

#"100. g H"_2"O"#

Explanation:

The problem essentially wants you to convert a given amount of energy to grams of steam at #100^@"C"#, so right from the start, you know that you're going to need a conversion factor.

This conversion factor is called the enthalpy of vaporization of water

#DeltaH_"vap" = "2257 J g"^(-1)#

http://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html

The enthalpy of vaporization tells you how much heat is needed in order to convert #"1 g"# of liquid water at its normal boiling point of #100^@"C"# to steam at #100^@"C"#.

In your case, you need #"2257 J"# of heat in order to convert #"1 g"# of liquid water at #100^@"C"# to steam at #100^@"C"#.

This means that #"226,000 J"# will allow you to convert

#"226,000" color(red)(cancel(color(black)("J"))) * "1 g"/(2257color(red)(cancel(color(black)("J")))) = color(darkgreen)(ul(color(black)("100. g")))#

The answer is rounded to three sig figs, the number of sig figs you have for the amount of heat.