Question #d2b4f

1 Answer
Oct 7, 2017

Answer:

The answer is indeed (B) #"203 g"#.

Explanation:

The first thing that you need to do here is to figure out how much heat is being given off when #"30.1 g"# of steam, presumably at #100^@"C"#, are being converted to #"30.1 g"# of liquid water at #100^@"C"#.

To do that, you can use the standard molar enthalpy of vaporization of water, #DeltaH_"vap"^@#, which tells you how much heat is being given off when #1# mole of water is converted from steam at #100^@"C"# to liquid water at #100^@"C"#, at standard pressure.

So, use the molar mass of water to convert the mass of steam to moles

#30.1 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.671 moles H"_2"O"#

You can thus say that condensing #"30.1 g"# of steam at #100^@"C"# to lqiuid water at #100^@"C"# will give off

#1.671 color(red)(cancel(color(black)("moles H"_2"O"))) * "40.7 kJ"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "68.01 kJ"#

Now, the specific heat of water tells you how much heat is needed to increase the temperature of #"1 g"# of water by #1^@"C"#.

More specifically, you know that in order to increase the temperature of #"1 g"# of water by #1^@"C"#, you need to provide the sample with #"4.184 J"# of heat.

In your case, you need to change the temperature of the sample by

#100.0^@"C" - 20.0^@"C" = 80.0^@"C"#

so start by calculating how much heat would be needed to do that for a given mass of water.

#80.0 color(red)(cancel(color(black)(""^@"C"))) * "4.184 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "334.72 J g"^(-1)#

This tells you that in order to increase the temperature of #"1 g"# of water by #80.0^@"C"#, you need to supply the sample with #"334.72 J"# of heat.

This means that the heat given off by the steam #-># liquid phase change will be enough to heat--remember that #"1 kJ" = 10^3# #"J"#

#68.01 * 10^3 color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(334.72 color(red)(cancel(color(black)("J")))))^(color(blue)("for an increase of 80.0"^@"C")) = color(darkgreen)(ul(color(black)("203 g")))#

The answer is rounded to three sig figs.