# 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

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

##### 1 Answer

I get

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

(1)Liquid water @#100^@ "C"# #stackrel("boil "" ")(->)# Water vapor @#100^@ "C"#

(2)Water vapor @#100^@ "C"# #stackrel("heat up"" ")(->)# Water vapor @#115^@ "C"#

The **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" (

#\mathbf(q_("vap",w) = n_wDelta"H"_("vap",w))# where:

#q_("vap",w)# is theheat flowinto the water that vaporizes it.#n_w# is the#\mathbf("mol")# sof water that were vaporized.#Delta"H"_("vap",w)# is theenthalpyfor 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")))#

#=# #color(green)("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

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.

#\mathbf(q_w = m_wc_wDeltaT_w = m_wc_w(T_(f,w) - T_(i,w)))# where:

#q_w# is theheat flowused to heat the water by#15^@ "C"# .#m_w# is themassof the water in#"g"# . Note that this had not changed after the water changed into a vapor/gas.#c_w = "1.841 J/g"cdot"K" = "1.841 J/g"cdot""^@ "C"# is thespecific heat capacityof watervapor. Water vapor has a lower specific heat capacity because its molecules are more active, and they release heat more easily through their faster motions.#DeltaT_w# is thechange in temperatureduring the heating process. This can be in either#""^@ "C"# or#"K"# , but I chose#0^@ "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"))#

#=# #"276.15 J"#

Don't forget to convert your heat units to match! We could use

**OVERALL HEAT REQUIRED**

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

#color(blue)(q_"tot") = q_("vap",w)^(100^@ "C") + q_w^(100^@ "C" -> 115^@ "C")#

#= 22.6 + 0.27615 = color(blue)("22.9 kJ")#