# Question #d2b4f

##### 1 Answer

The answer is indeed **(B)**

#### Explanation:

The first thing that you need to do here is to figure out how much heat is being **given off** when *liquid water* at

To do that, you can use the *standard molar enthalpy of vaporization of water*, **mole** of water is converted from *steam* at

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

#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

More specifically, you know that in order to increase the temperature of

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

This means that the heat given off by the steam

#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**.