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MCAT Physical : Thermodynamic Systems and Calorimetry

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Example Question #1 : Thermodynamic Systems And Calorimetry

A scientist prepares an experiment to demonstrate the second law of thermodynamics for a chemistry class. In order to conduct the experiment, the scientist brings the class outside in January and gathers a cup of water and a portable stove.

The temperature outside is –10 degrees Celsius. The scientist asks the students to consider the following when answering his questions:

Gibbs Free Energy Formula:

ΔG = ΔH – TΔS

Liquid-Solid Water Phase Change Reaction:

H₂O(l) ⇌ H₂O(s) + X

The scientist prepares two scenarios.

Scenario 1:

The scientist buries the cup of water outside in the snow, returns to the classroom with his class for one hour, and the class then checks on the cup. They find that the water has frozen in the cup.

Scenario 2:

The scientist then places the frozen cup of water on the stove and starts the gas. The class finds that the water melts quickly.

After the water melts, the scientist asks the students to consider two hypothetical scenarios as a thought experiment.

Scenario 3:

Once the liquid water at the end of scenario 2 melts completely, the scientist turns off the gas and monitors what happens to the water. Despite being in the cold air, the water never freezes.

Scenario 4:

The scientist takes the frozen water from the end of scenario 1, puts it on the active stove, and the water remains frozen.

In this situation described in the passage, which of the following is true?

Possible Answers:

The system is all of the universe, except the stove used to input energy.

The cup of water is the system, and the stove is the surroundings.

The cup of water is the system, and the snow is the surroundings.

The cup of water, snow, and stove comprise the system; the universe is the surroundings

The cup of water is the system, and the snow, stove, and remainder of the universe are the surroundings.

Correct answer:

The cup of water is the system, and the snow, stove, and remainder of the universe are the surroundings.

Explanation:

The Gibbs Free Energy equation makes use of clearly delineated systems and surroundings. In this example, the water is freezing or melting depending on conditions. This is accompanied by thermal exchanges with other players, such as the snow and stove. Thus, the water is the system, and everything else (technically, everything else in the universe) comprises the surroundings.

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Example Question #2 : Thermodynamic Systems And Calorimetry

The combustion of propane is given by the following formula.

$C_{3}H_{8}(g) + 4O_{2}(g) \rightarrow 2CO(g) + CO_{2}(g) + 4H_{2}O(g)$

If the heats of formation for CO, CO₂, H₂O, and C₃H₈ are -110.5 kJ/mol, -393.5 kJ/mol, -241.8 kJ/mol, and -103.85 kJ/mol, respectively, what is the heat of reaction for the combustion of propane?

Possible Answers:

641.95kJ

1477.9kJ

–1477.9kJ

–641.95kJ

Correct answer:

–1477.9kJ

Explanation:

Given the fact that combustion reactions are exothermic, you should expect the heat of reaction to be negative (ruling out two answer choices). The heat of reaction is equal to the heat of formation of the products minus the heat of formation of the reactants. Be sure to refer to the balanced equation for the correct number of moles for each compound.

$H_{rxn}=(H_f\ products)-(H_f\ reactants)$

$H_f\ products=(2mol)(-110.5\frac{kJ}{mol})+(1mol)(-393.5\frac{kJ}{mol})+(4mol)(-241.8\frac{kJ}{mol})$
$H_f\ products= -1581.7kJ$

$H_f\ reactants=(1mol)(-103.85\frac{kJ}{mol})=-103.85kJ$

$H_{rxn}=-1581.7kJ-(-103.85kJ)=-1477.9kJ$

Oxygen is not included, as it is in elemental form and therefore has a heat of formation equal to zero.

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Example Question #1 : Thermodynamic Systems And Calorimetry

The combustion of liquid hexane in air at 298K gives gaseous carbon dioxide and liquid water, as shown in this reaction.

$2C_{6}H_{14}(l) + 19O_{2}(g)\rightarrow 12CO_{2}(g) + 14H_{2}O(l)$

$\Delta H_{f}^{0}$ of $C_6H_{14}(l)$ is $-167.4 \frac{kJ}{mol}$ .

$\Delta H_{f}^{0}$ of CO_2(g) is $-393.5 \frac{kJ}{mol}$ .

$\Delta H_{f}^{0}$ of H_2O(l) is $-285.8 \frac{kJ}{mol}$ .

Calculate the $\Delta H_{rxn}^{0}$ for the combustion of hexane liquid hexane at 298K.

Possible Answers:

$9058.6\frac{kJ}{mol}$

$8384.2\frac{kJ}{mol}$

$-9058.6\frac{kJ}{mol}$

$-8384.2\frac{kJ}{mol}$

$-4194.2\frac{kJ}{mol}$

Correct answer:

$-8384.2\frac{kJ}{mol}$

Explanation:

To calculate the $\Delta H_{f}^{0}$ , the following formula is used. Remember that the coefficients of the balanced chemical equation must be included, as shown. Also, recall that the $\Delta H_{f}^{0}$ of any pure element is zero.

$\Delta H_{f}^{0} = \sum \Delta H_{products}^{0} - \sum \Delta H_{reactants}^{0}$

$\Delta H_{f}^{0} = 12\Delta H_{f}^{0} (CO_{2}) + 14\Delta H_{f}^{0} (H_{2}O)-(2\Delta H_{f}^{0}(C_{6}H_{14}) + 19\Delta H_{f}^{0} (O_{2}))$

Now we can plug in the given values and solve for the enthalpy of reaction.

$\Delta H_{f}^{0} = 12(-393.5\frac{kJ}{mol})+14(-285.8\frac{kJ}{mol})- (2(-167.4\frac{kJ}{mol})+19(0\frac{kJ}{mol}))$

$\Delta H_{f}^{0} = -4722\frac{kJ}{mol}+(-3997\frac{kJ}{mol})- (-334.8\frac{kJ}{mol}+0\frac{kJ}{mol})$

$\Delta H_{f}^{0} = -8384.2\frac{kJ}{mol}$

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Example Question #4 : Thermodynamic Systems And Calorimetry

How much energy must a stove transfer to completely transform $\small 1g$ of water at $\small 0^oC$ into steam?

$\Delta H_{vap\ water}=2.26\frac{kJ}{g}$

$\small c_{water}=4.148\frac{J}{g ^oC}$

Possible Answers:

2405J

2.26J

2675J

205.4J

9.56J

Correct answer:

2675J

Explanation:

To turn the water into steam, the stove must first raise the temperature of the water, and then provide energy to change the phase. This requires two distinct steps.

Find the energy to raise the temperature of the water using the equation $q=mc\Delta T$ . We know our mass, the specific heat of water, and the change in temperature. Use the given values to find the necessary heat.

$q=(1g)(4.148\frac{J}{g^oC})(100^oC-0^oC)$

q=414.8J

Now, we need to find the energy needed to convert the water to a gas. We will use the equation $q=m\Delta H_{vap}$ .

$q=(1g)(2.26\frac{kJ}{g})=2.26kJ$

Finally, we add the energy for the two steps to find the total energy required.

414.8J+2260J=2674.8J

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Example Question #5 : Thermodynamic Systems And Calorimetry

$2S + 3O_{2}\rightarrow2SO_{3}$ $\Delta H = -791.4 kJ$

In the above reaction, how much heat will be released if 74.0g of sulfur reacts with excess oxygen? Round to the nearest 10kJ.

Possible Answers:

460kJ

1370kJ

790kJ

390kJ

910kJ

Correct answer:

910kJ

Explanation:

Basically, this is a unit conversion problem. Starting with grams of sulfur, convert to moles of sulfur, and finally to kJ of heat released. Note that the given enthalpy of reaction, $\Delta H$ , is the amount of heat released when two moles of sulfur react with three moles of oxygen.

$\small \left ( \frac{74.0 g S}{1} \right )\left ( \frac{1 mol S}{32.06 g S} \right )\left ( \frac{-791.4kJ}{2 mol S} \right )=-913.34kJ\approx-910kJ$

Note that a negative enthalpy of reaction means the process is exothermic and releases heat, while a positive enthalpy of reaction would mean the process is endothermic and absorbs heat.

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Example Question #6 : Thermodynamic Systems And Calorimetry

A 50g sample of an unknown substance is heated to 100^oC in a tub of boiling water. It is then quickly removed and placed into an insulated jar holding 200mL of water, initially at 20^oC. The final equilbrium temperature of the system is 30^oC. Approximately, what is the specific heat of this unknown substance?

Specific heat of water is 4.187J/g^oC.

Density of water is 1g/mL.

Possible Answers:

$5.0 J/g^{o}C$

$1.5 J/g^{o}C$

$2.4 J/g^{o}C$

$1.2 J/g^{o}C$

$3.3 J/g^{o}C$

Correct answer:

$2.4 J/g^{o}C$

Explanation:

Since the system is isolated, the amount of heat transferred away from the unknown substance must equal the heat transferred to the water. To calculate these heats, use $\Delta Q = mc\Delta T$

For the water, $\Delta Q = (200g)(4.187 J/g^{o}C)(10^{o}C) = 8374 J$ .

For the unknown substance, $\Delta Q = (50g)c_x(70^{o}C) = 3500c_xJ$

Set these values equal to get c_x = 2.39 J/g^oC, approximately 2.4 J/g^oC.

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Example Question #7 : Thermodynamic Systems And Calorimetry

$\small L_{f} = 334 \frac{J}{g}$

$\small L_{v}=2260 \frac{J}{g}$

$\small c_{water}=4.186\frac{J}{g*C}$

$\small c_{ice}=2.05\frac{J}{g*C}$

How much energy is needed to change a 50g ice cube at -30^oC into 50g of water at 40^oC? Use the above quantities as needed, and round the answer to the nearest kJ.

Possible Answers:

17kJ

28kJ

7kJ

11kJ

22kJ

Correct answer:

28kJ

Explanation:

There are three distinct steps in this transformation: 1) heating the ice from -30^oC to 0^oC, 2) melting the ice at 0^oC, and 3) heating the water from 0^oC to 40^oC.

When there is a temperature change, we use $\small \Delta Q = mc \Delta T$ , and when there is melting or freezing, we use $\small \Delta Q = mL_{f}$ .

1) $\small \small \Delta Q = mc_{ice}\Delta T = (50g)(2.05\frac{J}{g*C})(30C)=3075J$

2) $\small \small \Delta Q = mL_{f} = (50g)(344 \frac{J}{g})=16700J$

3) $\small \Delta Q = mc_{water}\Delta T = (50g)(4.186\frac{J}{g*C})(40C) =8372 J$

Adding these three pieces together gives the total enthalpy change, which is equal to change in energy.

$\small 3075 + 16700 + 8372 = 28147 J\approx28kJ$

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Example Question #1 : Thermodynamic Systems And Calorimetry

Which term correctly identifies a system in which there is an exchange of heat with the surrounding environment, but no exchange of matter?

Possible Answers:

Conserved system

Isolated system

Closed system

Open system

Thermodynamic system

Correct answer:

Closed system

Explanation:

In a closed system, energy and work can be exchanged with the surrounding area but matter is contained. Consider, for example, a beaker of ice. Heat from the surroundings it transferred to the ice, causing it to melt, but there is no exchange of matter. No external compounds react with the ice and none of the ice leaves the beaker.

In an open system, both heat and matter can be exchanged. In an isolated system neither heat, nor matter is exchanged.

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Example Question #1 : Thermodynamic Systems And Calorimetry

$c_{ice}=2.03\frac{J}{g^oC}$

$c_{water}=4.18\frac{J}{g^oC}$

$\Delta H_{fus}=334\frac{J}{g}$

You begin an experiment with $\small 12.0g$ of ice at $\small -3^oC$ . How much heat is required to melt the ice and raise its temperature to $\samll 15^oC$ ?

Possible Answers:

1380.7J

0.825J

4833.48J

48.3kJ

419J

Correct answer:

4833.48J

Explanation:

This process can be divided into three stages: the raising of the temperature of the ice, the melting of the ice into water, and the raising of the temperature of the water. The first and third step use the specific heat of the substance to calculate the heat required, while the second step uses the heat of fusion.

Step 1 and 3: $q=mc\Delta T$

Step 2: $q=m\Delta H_{fus}$

We know the mass of the water, the change in temperature, and the constant values. Using these values, we can sum the heat required for each step to get our final answer.

$q_{total}=q_1+q_2+q_3$

$q_1=(12.0g)(2.03\frac{J}{g^oC})(0^0C-(-3^oC))=73.08J$

$q_2=(12.0g)(334\frac{J}{g})=4008J$

$q_3=(12.0g)(4.18\frac{J}{g^oC})(15^oC-0^oC)=752.4J$

$q_{total}=73.08J+4008J+752.4J=4833.48J$

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Example Question #8 : General Thermodynamics

Use the following values for water as needed.

$\Delta H^o_{fus}=334\frac{J}{g}$

$\Delta H^o_{vap}=2260\frac{J}{g}$

$c=4.186\frac{J}{g^oC}$

$\rho = 1\frac{g}{mL}$

If burning wood releases $\small 20kJ$ of heat energy per gram of wood consumed, what mass of wood must be consumed to heat $\small 1L$ of water from $\small 20^oC$ to $\small 100^oC$ , and then to convert it to water vapor?

Possible Answers:

17g

335g

134g

130g

2595g

Correct answer:

130g

Explanation:

There are two processes requiring added heat in this problem:

1. Raising the temperature of the liquid water from $\small 20^oC$ to $\small 100^oC$ (use $\Delta Q = mc\Delta T$ )

2. Boiling the water at a constant temperature of $\small 100^oC$ (use $\Delta Q = m\Delta H^o_{vap}$ )

To use either of these equation, we need to find the mass of the water using the relation between mass, density, and volume.

$m = \rho*V= 1\frac{g}{mL}*1000mL=1000g$

Use this mass with the given specific heat and temperatures to find the heat for part 1 of the process.

$\Delta Q = (1000g)(4.186\frac{J}{g^oC})(100^oC - 20^oC)=334,880J$

Then, use the mass with the given heat of vaporization to find the energy needed to convert the water to water vapor.

$\Delta Q =(1000g) (2260\frac{J}{g})= 2,260,000J$

Sum the energies for step 1 and step 2.

$334,880 J + 2,260,000 J = 2,594,880J\approx 2,595kJ$

This is the total amount of energy needed from the burning wood. Use stoichiometry to find the grams of wood needed to produce this amount of energy.

$2595kJ*(\frac{1 g}{20kJ})=129.7g\approx 130g$

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MCAT Physical : Thermodynamic Systems and Calorimetry

Study concepts, example questions & explanations for MCAT Physical

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Example Question #1 : Thermodynamic Systems And Calorimetry

Example Question #2 : Thermodynamic Systems And Calorimetry

Example Question #1 : Thermodynamic Systems And Calorimetry

Example Question #4 : Thermodynamic Systems And Calorimetry

Example Question #5 : Thermodynamic Systems And Calorimetry

Example Question #6 : Thermodynamic Systems And Calorimetry

Example Question #7 : Thermodynamic Systems And Calorimetry

Example Question #1 : Thermodynamic Systems And Calorimetry

Example Question #1 : Thermodynamic Systems And Calorimetry

Example Question #8 : General Thermodynamics

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