# Question d2b4f

Oct 7, 2017

The answer is indeed (B) $\text{203 g}$.

#### Explanation:

The first thing that you need to do here is to figure out how much heat is being given off when $\text{30.1 g}$ of steam, presumably at ${100}^{\circ} \text{C}$, are being converted to $\text{30.1 g}$ of liquid water at ${100}^{\circ} \text{C}$.

To do that, you can use the standard molar enthalpy of vaporization of water, $\Delta {H}_{\text{vap}}^{\circ}$, which tells you how much heat is being given off when $1$ mole of water is converted from steam at ${100}^{\circ} \text{C}$ to liquid water at ${100}^{\circ} \text{C}$, 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 $\text{30.1 g}$ of steam at ${100}^{\circ} \text{C}$ to lqiuid water at ${100}^{\circ} \text{C}$ 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 $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

More specifically, you know that in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, you need to provide the sample with $\text{4.184 J}$ of heat.

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

${100.0}^{\circ} \text{C" - 20.0^@"C" = 80.0^@"C}$

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 $\text{1 g}$ of water by ${80.0}^{\circ} \text{C}$, you need to supply the sample with $\text{334.72 J}$ of heat.

This means that the heat given off by the steam $\to$ liquid phase change will be enough to heat--remember that $\text{1 kJ} = {10}^{3}$ $\text{J}$

$68.01 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))) * overbrace("1 g"/(334.72 color(red)(cancel(color(black)("J")))))^(color(blue)("for an increase of 80.0"^@"C")) = color(darkgreen)(ul(color(black)("203 g}}}}$

The answer is rounded to three sig figs.