Question

Question #d2b4f

Answer 1

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.