# Is the answer to this question 221 J? If not, can someone explain it to me?

##### 1 Answer

No, not even close.

#### Explanation:

**!! VERY LONG ANSWER !!**

The problem wants you to calculate the total change in enthalpy, **ice** at **steam** at

The first thing to do here is figure out the *sign* of **provide heat**.

In other words, you're dealing with a series of **endothermic processes**. The change in enthalpy associated with an endothermic process is **positive**, so right from the start you know that the

Now, in order to be able to do the actual calculations, you need to know

#c_"ice" = "2.06 J g"^(-1)""^@"C"^(-1)#

#DeltaH_"fus" = "333.55 J g"^(-1)#

#c_"water" = "4.184 J g"^(-1)""^@"C"^(-1)#

#DeltaH_"vap" = "2259 J g"^(-1)#

#c_"steam" = "2.02 J g"^(-1)""^@"C"^(-1)#

So, break up your process into **five** parts

*Going from***ice**at#-10.0^@"C"# *to***ice**at#0^@"C"#

The heat required to heat your sample of ice from

#q_1 = m_"ice" * c_"ice" * DeltaT_"ice"#

Plug in your values to find

#q_1 = 21.5 color(red)(cancel(color(black)("g"))) * "2.06 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * [0 - (-10.0)]color(red)(cancel(color(black)(""^@"C")))#

#q_1 = "442.9 J"#

*Going from***ice**at#0^@"C"# *to***liquid water**at#0^@"C"#

This is where you encounter the first **phase change**. All the heat that you provide to the sample of **ice** at *stuck in place* and convert it to **liquid water** at

Phase changes occur at **constant temperature**, which is why your phase diagram shows two *horizontal lines*

one is marked with

#DeltaH_"melt" -># here is where melting / freezing take placeone marked with

#DeltaH_"vap" -># here is where boiling / condensation take place

The heat required to promote a solid

#q_2 = m_"ice" * DeltaH_"fus"#

Plug in your values to find

#q_2 = 21.5 color(red)(cancel(color(black)("g"))) * "333.55 J" color(red)(cancel(color(black)("g"^(-1))))#

#q_2 = "7171.3 J"#

*Going from***liquid water**at#0^@"C"# *to***liquid water**at#100^@"C"#

The heat required to heat your sample from liquid water at

#q_3 = M_"water" * c_"water" * DeltaT_"water"#

Plug in your values to find

#q_3 = 21.5 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (100 - 0)color(red)(cancel(color(black)(""^@"C")))#

#q_3 = "8995.6 J"#

*Going from***liquid water**at#100^@"C"# *to***steam**at#100^@"C"#

This is where you encounter the second **phase change**. All the heat you supply to the sample of liquid water will be used to help the water molecules **completely overpower** the hydrogen bonds and start going into the gaseous state.

The heat required to promote a liquid

#q_4 = m_"water" * DeltaH_"vap"#

Plug in your values to find

#q_4 = 21.5 color(red)(cancel(color(black)("g"))) * "2259 J" color(red)(cancel(color(black)("g"^(-1))))#

#q_4 = "48568.5 J"#

*Going from***steam**at#100^@"C"# *to***steam**at#115.0^@"C"#

Finally, you can calculate the heat required to heat your sample of steam from

#q_5 = m_"steam" * c_"steam" * DeltaT_"steam"#

Plug in your values to find

#q_5 = 21.5 color(red)(cancel(color(black)("g"))) * "2.02 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (115.0 - 100)color(red)(cancel(color(black)(""^@"C")))#

#q_5 = "651.5 J"#

Now, the **total heat** required to get your sample from ice at

#q_"total" = q_1 + q_2 + q_3 + q_4 + q_5#

In your case, you have

#q_"total" = "442.9 J" + "7171.3 J" + "8995.6 J" + "48568.5 J" + "651.5 J"#

#q_"total" = "65829.8 J"#

The problem wants you to find the change in enthalpy **per mole**, os use water's **molar mass** to figure out how many moles of water you have in your sample

#21.5 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.1934 moles H"_2"O"#

You can thus say that the heat required to convert **one mole** of ice at

#1 color(red)(cancel(color(black)("mole H"_2"O"))) * "65829.8 J"/(1.1934color(red)(cancel(color(black)("moles H"_2"O")))) = "55161.6 J"#

Expressed in *kilojoules* and rounded to three **sig figs**, the amoiunt of heat needed will be

#55161.6 color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = "55.2 kJ"#

Since **one mole** of water requires

#color(green)(|bar(ul(color(white)(a/a)color(black)(DeltaH = +"55.2 kJ mol"^(-1))color(white)(a/a)|)))#