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

## Aug 2, 2016

No, not even close.

#### Explanation:

The problem wants you to calculate the total change in enthalpy, $\Delta H$, that accompanies the conversion of $\text{21.5 g}$ of ice at $- {10.0}^{\circ} \text{C}$ to steam at ${11.50}^{\circ} \text{C}$.

The first thing to do here is figure out the sign of $\Delta H$. In order to melt ice, heat liquid water, boil liquid water, and heat steam, you must 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 $\Delta H$ will be a positive value.

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

${c}_{\text{ice" = "2.06 J g"^(-1)""^@"C}}^{- 1}$

$\Delta {H}_{\text{fus" = "333.55 J g}}^{- 1}$

${c}_{\text{water" = "4.184 J g"^(-1)""^@"C}}^{- 1}$

$\Delta {H}_{\text{vap" = "2259 J g}}^{- 1}$

${c}_{\text{steam" = "2.02 J g"^(-1)""^@"C}}^{- 1}$

So, break up your process into five parts

$\textcolor{w h i t e}{a}$

• Going from ice at $- {10.0}^{\circ} \text{C}$ to ice at ${0}^{\circ} \text{C}$

The heat required to heat your sample of ice from $- {10.0}^{\circ} \text{C}$ to ${0}^{\circ} \text{C}$ can be calculated using the equation

${q}_{1} = {m}_{\text{ice" * c_"ice" * DeltaT_"ice}}$

Plug in your values to find

${q}_{1} = 21.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "2.06 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * [0 - (-10.0)]color(red)(cancel(color(black)(""^@"C}}}}$

${q}_{1} = \text{442.9 J}$

$\textcolor{w h i t e}{a}$

• Going from ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$

This is where you encounter the first phase change. All the heat that you provide to the sample of ice at ${0}^{\circ} \text{C}$ will be used to help the molecules overpower the hydrogen bonds that are keeping them stuck in place and convert it to liquid water at 0^@"C.

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

• one is marked with $\Delta {H}_{\text{melt}} \to$ here is where melting / freezing take place

• one marked with $\Delta {H}_{\text{vap}} \to$ here is where boiling / condensation take place

The heat required to promote a solid $\to$ liquid phase change can be calculated using the equation

${q}_{2} = {m}_{\text{ice" * DeltaH_"fus}}$

Plug in your values to find

${q}_{2} = 21.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g"))) * "333.55 J" color(red)(cancel(color(black)("g}}^{- 1}}}}$

${q}_{2} = \text{7171.3 J}$

$\textcolor{w h i t e}{a}$

• Going from liquid water at ${0}^{\circ} \text{C}$ to liquid water at ${100}^{\circ} \text{C}$

The heat required to heat your sample from liquid water at ${0}^{\circ} \text{C}$ to liquid water at ${100}^{\circ} \text{C}$ can be calculated using the equation

${q}_{3} = {M}_{\text{water" * c_"water" * DeltaT_"water}}$

Plug in your values to find

${q}_{3} = 21.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (100 - 0)color(red)(cancel(color(black)(""^@"C}}}}$

${q}_{3} = \text{8995.6 J}$

$\textcolor{w h i t e}{a}$

• Going from liquid water at ${100}^{\circ} \text{C}$ to steam at ${100}^{\circ} \text{C}$

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 $\to$ vapor phase change can be calculated using the equation

${q}_{4} = {m}_{\text{water" * DeltaH_"vap}}$

Plug in your values to find

${q}_{4} = 21.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g"))) * "2259 J" color(red)(cancel(color(black)("g}}^{- 1}}}}$

${q}_{4} = \text{48568.5 J}$

$\textcolor{w h i t e}{a}$

• Going from steam at ${100}^{\circ} \text{C}$ to steam at ${115.0}^{\circ} \text{C}$

Finally, you can calculate the heat required to heat your sample of steam from ${100}^{\circ} \text{C}$ to ${115.0}^{\circ} \text{C}$ by using the equation

${q}_{5} = {m}_{\text{steam" * c_"steam" * DeltaT_"steam}}$

Plug in your values to find

${q}_{5} = 21.5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "2.02 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (115.0 - 100)color(red)(cancel(color(black)(""^@"C}}}}$

${q}_{5} = \text{651.5 J}$

$\textcolor{w h i t e}{a}$

Now, the total heat required to get your sample from ice at $- {10.0}^{\circ} \text{C}$ to steam at ${115.0}^{\circ} \text{C}$ will be equal to

${q}_{\text{total}} = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5}$

${q}_{\text{total" = "442.9 J" + "7171.3 J" + "8995.6 J" + "48568.5 J" + "651.5 J}}$

${q}_{\text{total" = "65829.8 J}}$

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 $- {10.0}^{\circ} \text{C}$ to steam at ${115.0}^{\circ} \text{C}$ is equal to

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 $\text{55.2 kJ}$ of heat in order to go from ice at $- {10.0}^{\circ} \text{C}$ to steam at ${115.0}^{\circ} \text{C}$, it follows htat the change in enthalpy for the overall process is

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\Delta H = + {\text{55.2 kJ mol}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$