# Use the following data to calculate the energy required to convert 10.0 g of liquid water at 100.0C to steam at 115.0C under a constant 1 atm pressure.?

## Use the following data to calculate the energy required to convert 10.0 g of liquid water at 100.0C to steam at 115.0C under a constant 1 atm pressure.? Specific heat of liquid water: 4.184 J/(g K) Specific heat of water vapor: 1.841 J/(g K) Heat of vaporization for water: 40.67 kJ/mol

May 18, 2016

I get $\text{22.9 kJ}$.

It helps to outline the path you go through first, so you can organize your answer.

(1) Liquid water @ ${100}^{\circ} \text{C}$ $\stackrel{\text{boil "" }}{\to}$ Water vapor @ ${100}^{\circ} \text{C}$

(2) Water vapor @ ${100}^{\circ} \text{C}$ $\stackrel{\text{heat up"" }}{\to}$ Water vapor @ ${115}^{\circ} \text{C}$

The $\text{1 atm}$ pressure is just for context, and says you are at everyday conditions. It also says you are at constant pressure, which, wouldn't you know it, is a hint!

(1)

When boiling water, it goes through a vapor-pressure-atmospheric pressure equilibrium where all the heat used goes into vaporizing the water.

This agrees with the condition that the temperature does not change on this step.

For this, we use the concept of a "latent heat of vaporization" (Delta"H"_("vap",w)) in the following equation:

$\setminus m a t h b f \left({q}_{\text{vap",w) = n_wDelta"H"_("vap} , w}\right)$

where:

• ${q}_{\text{vap} , w}$ is the heat flow into the water that vaporizes it.
• ${n}_{w}$ is the $\setminus m a t h b f \left(\text{mol}\right)$s of water that were vaporized.
• Delta"H"_("vap",w) is the enthalpy for the vaporization process.

As a side note, if you recall from the premises of the question, this is a constant-pressure condition, uncoincidentally very similar to whenever you were to calculate the heat flow in a standard reaction from this equation:

Delta"H"_"rxn" = (q_"rxn")/("mols limiting reagent")

Anyways, the heat flow for part (1) is:

color(green)(q_("vap",w)) = stackrel(n_w)overbrace((10.0 cancel("g H"_2"O"))(cancel(("1 mol H"_2"O"))/("18.015" cancel("g H"_2"O"))))stackrel(Delta"H"_("vap",w))overbrace(("40.67 kJ/"cancel("mol")))

$=$ $\textcolor{g r e e n}{\text{22.6 kJ}}$

Turns out we didn't need the specific heat capacity of liquid water.

That's because we already have the enthalpy, which already gives you the proper units of $\text{kJ}$ in the end.

Also, if you had used the specific heat capacity of liquid water, you would have had a temperature dependence when there is none, as we stated that the temperature was constant.

(2)

Now you're just heating water vapor without any phase changes, so you will be utilizing the specific heat capacity of water vapor.

$\setminus m a t h b f \left({q}_{w} = {m}_{w} {c}_{w} \Delta {T}_{w} = {m}_{w} {c}_{w} \left({T}_{f , w} - {T}_{i , w}\right)\right)$

where:

• ${q}_{w}$ is the heat flow used to heat the water by ${15}^{\circ} \text{C}$.
• ${m}_{w}$ is the mass of the water in $\text{g}$. Note that this had not changed after the water changed into a vapor/gas.
• ${c}_{w} = \text{1.841 J/g"cdot"K" = "1.841 J/g"cdot""^@ "C}$ is the specific heat capacity of water vapor. Water vapor has a lower specific heat capacity because its molecules are more active, and they release heat more easily through their faster motions.
• $\Delta {T}_{w}$ is the change in temperature during the heating process. This can be in either $\text{^@ "C}$ or $\text{K}$, but I chose ${0}^{\circ} \text{C}$ to limit the number of required conversions since celsius and kelvin intervals are identical.

So, we just get:

color(green)(q_w) = ("10.0" cancel("g H"_2"O"))("1.841 J/"cancel"g"cdotcancel(""^@ "C"))(115^@ cancel("C") - 100^@ cancel("C"))

$=$ $\text{276.15 J}$

Don't forget to convert your heat units to match! We could use $\text{kJ}$, then, so you would get "276.15" cancel("J") xx "1 kJ"/(1000 cancel("J")) = color(green)("0.27615 kJ").

OVERALL HEAT REQUIRED

To get the overall heat, simply sum these to get:

$\textcolor{b l u e}{{q}_{\text{tot") = q_("vap",w)^(100^@ "C") + q_w^(100^@ "C" -> 115^@ "C}}}$

$= 22.6 + 0.27615 = \textcolor{b l u e}{\text{22.9 kJ}}$