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# 17.12: Multi-Step Problems with Changes of State

Difficulty Level: At Grade Created by: CK-12
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Practice Multi-Step Problems with Changes of State
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#### Which takes more heat – melting or boiling?

You have a cube of ice. Which process will take more energy – the melting of that ice cube or the conversion of the water to steam? The short answer is that more energy is needed to convert the water to steam. The long answer is really a question: how do you get from one point to the other? What is the temperature of the ice? What is the mass of that ice cube? A lot goes into taking the material from the starting point to the end-point.

### Multi-Step Problems with Changes of State

Heating curves show the phase changes that a substance undergoes as heat is continuously absorbed.

Heating curve of water.

The specific heat of a substance allows us to calculate the heat absorbed or released as the temperature of the substance changes. It is possible to combine that type of problem with a change of state to solve a problem involving multiple steps. Figure above shows ice at -30°C being converted in a five-step process to gaseous water (steam) at 140°C. It is now possible to calculate the heat absorbed during that entire process. The process and the required calculation is summarized below.

1. Ice is heated from -30°C to 0°C. The heat absorbed is calculated by using the specific heat of ice and the equation ΔH=cp×m×ΔT\begin{align*}\Delta H = c_p \times m \times \Delta T\end{align*}.
2. Ice is melted at 0°C. The heat absorbed is calculated by multiplying the moles of ice by the molar heat of fusion.
3. Water at 0°C is heated to 100°C. The heat absorbed is calculated by using the specific heat of water and the equation ΔH=cp×m×ΔT\begin{align*}\Delta H = c_p \times m \times \Delta T\end{align*}.
4. Water is vaporized to steam at 100°C. The heat absorbed is calculated by multiplying the moles of water by the molar heat of vaporization.
5. Steam is heated from 100°C to 140°C. The heat absorbed is calculated by using the specific heat of steam and the equation ΔH=cp×m×ΔT\begin{align*}\Delta H = c_p \times m \times \Delta T\end{align*}.

#### Sample Problem: Multi-Step Problems using a Heating Curve

Calculate the total amount of heat absorbed (in kJ) when 2.00 mol of ice at -30.0°C is converted to steam at 140.0°C. The required specific heats can be found in the table in "Heat Capacity and Specific Heat".

Step 1: List the known quantities and plan the problem.

Known

• 2.00 mol ice = 36.04 g ice
• cp(ice)=2.06 J/gC\begin{align*}c_p (\text{ice}) = 2.06 \text{ J/g}^\circ \text{C}\end{align*}
• cp(water)=4.18 J/gC\begin{align*}c_p(\text{water}) = 4.18 \text{ J/g}^\circ \text{C}\end{align*}
• cp(steam)=1.87 J/gC\begin{align*}c_ p(\text{steam}) = 1.87 \text{ J/g}^\circ \text{C}\end{align*}
• ΔHfus=6.01 kJ/mol\begin{align*}\Delta H_{\text{fus}} = 6.01 \text{ kJ/mol}\end{align*}
• ΔHvap=40.7 kJ/mol\begin{align*}\Delta H_{\text{vap}}= 40.7 \text{ kJ/mol}\end{align*}

Unknown

• ΔHtotal=? kJ\begin{align*}\Delta H_{\text{total}}= ? \text{ kJ}\end{align*}

Follow the steps previously described. Note that the mass of the water is needed for the calculations that involve the specific heat, while the moles of water is needed for the calculations that involve changes of state. All heat quantities must be in kilojoules so that they can be added together to get a total for the five-step process.

Step 2: Solve.

1. ΔH1=2.06 J/gC×36.04 g×30C×1 kJ1000 J=2.23 kJ\begin{align*}\Delta H_1=2.06 \text{ J/g}^\circ \text{C} \times 36.04 \text{ g} \times 30^\circ \text{C} \times \frac{1 \text{ kJ}}{1000 \text{ J}}=2.23 \text{ kJ}\end{align*}
2. ΔH2=2.00 mol×6.01 kJ1 mol=12.0 kJ\begin{align*}\Delta H_2=2.00 \text{ mol} \times \frac{6.01 \text{ kJ}}{1 \text{ mol}}=12.0 \text{ kJ}\end{align*}
3. ΔH3=4.18 J/gC×36.04 g×100C×1 kJ1000 J=15.1 kJ\begin{align*}\Delta H_3=4.18 \text{ J/g}^\circ \text{C} \times 36.04 \text{ g} \times 100^\circ \text{C} \times \frac{1 \text{ kJ}}{1000 \text{ J}}=15.1 \text{ kJ}\end{align*}
4. ΔH4=2.00 mol×40.7 kJ1 mol=81.4 kJ\begin{align*}\Delta H_4=2.00 \text{ mol} \times \frac{40.7 \text{ kJ}}{1 \text{ mol}}=81.4 \text{ kJ}\end{align*}
5. ΔH5=1.87 J/gC×36.04 g×40C×1 kJ1000 J=2.70 kJ\begin{align*}\Delta H_5=1.87 \text{ J/g}^\circ \text{C} \times 36.04 \text{ g} \times 40^\circ \text{C} \times \frac{1 \text{ kJ}}{1000 \text{ J}}=2.70 \text{ kJ} \end{align*}

ΔHtotal=ΔH1+ΔH2+ΔH3+ΔH4+ΔH5=113.4 kJ\begin{align*}\Delta H_{\text{total}}= \Delta H_1+ \Delta H_2+\Delta H_3+ \Delta H_4+ \Delta H_5=113.4 \text{ kJ}\end{align*}

The total heat absorbed as the ice at -30°C is heated to steam at 140°C is 113.4 kJ. The largest absorption of heat comes during the vaporization of the liquid water.

### Review

1. Why are two different sets of units used?
2. What other units problem do you need to be aware of?
3. What would you need to know to do calculations like this for acetone?

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