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AP Chemistry : Thermodynamics

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Example Questions

Example Question #1 : Thermodynamics

The following is a list of specific heat capacities for a few metals.

$\small C_{copper} = 0.385$

$\small C_{iron} = 0.444$

$\small C_{silver} = 0.240$

$\small C_{aluminum} = 0.900$

A 50g sample of an unknown metal is heated with 800 joules. If the temperature of the metal increases by 41.6^oC, what is the identity of the unknown metal?

Possible Answers:

Iron

Aluminum

Copper

Silver

Correct answer:

Copper

Explanation:

We need to find the specific heat of the unknown sample of metal in order to locate it on the list. We can do this by using the equation that allows us to determine the specific heat capacity of an element.

$\small q = mC\Delta T$

Since we know the change in temperature, we can simply plug in the values and solve for the value of $\small C$ .

$\small 800J = (50g)C(41.6^oC)$

$\small C = 0.385$

Going back to the list, we see that this is the specific heat capacity for copper, so we confirm that the unknown metal is copper.

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Example Question #1 : Calorimetry, Specific Heat, And Calculations

In which instance would a bomb calorimeter be more useful than a coffee-cup calorimeter?

Possible Answers:

None of the above

When measuring the mass of a substance

When measuring the heat capacity of a solid

When measuring the specific heat of water vapor

Correct answer:

When measuring the specific heat of water vapor

Explanation:

Bomb calorimeters are most useful when dealing with a gas, because they can operate well at high pressures. Coffee-cup calorimeters are not useful when water begins to boil, producing vapor.

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Example Question #1 : Calorimetry, Specific Heat, And Calculations

How much heat does it take to heat 100g ice at 0C to boiling point?

C_ice= 2.1 J/g^oC

C_water= 4.2 J/g^oC

ΔH_va_p= 2260 J/g

ΔH_fus=334 J/g

Possible Answers:

300.5 kJ

33.4 kJ

55.4 kJ

42 kJ

75.4 kJ

Correct answer:

75.4 kJ

Explanation:

You need heat for the phase change, using the enthalpy of fusion (100g*334 J/g = 33400 J). Add to this the heat to get to boiling point using the specific heat of water (100g*100C*4.2 J/g^oC = 42000 J). Totalling 75400 J (75.4 kJ)

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Example Question #2 : Calorimetry, Specific Heat, And Calculations

The specific heat capacity of an unknown liquid is $0.32\frac{J}{kg\cdot K}$ . The density of the liquid is $0.0321 \frac{g}{mL}$ If a chemist applies 243 J of heat to 300 mL of this liquid starting at $27.1^{\circ}C$ , what is the final temperature?

Possible Answers:

$78882\hspace{1 mm}K$

None of the available answers

$78882^{\circ}C$

$270.1^{\circ}C$

$2700.1^{\circ}C$

Correct answer:

$78882^{\circ}C$

Explanation:

First we will determine the mass of the liquid:

$300\hspace{1 mm}mL\times\frac{0.0321\hspace{1 mm}g}{1\hspace{1 mm}mL}=9.63\hspace{1 mm}g$

Now we will examine the relationship between heat and specific heat capacity:

$Q=cm\Delta T$

Where $Q$ is heat in Joules, $c$ is the specific heat capacity, $m$ is the mass and $\Delta T$ is the change in temperature. We can rearrange this

$\Delta T=\frac{Q}{cm}$

$\Delta T=\frac{243\hspace{1 mm}J}{0.32J\cdot kg^{-1}\cdot K^{-1}0.00963kg}=78855\hspace{1 mm}K$

If we begin at $27.1^{\circ}C$ , we will end at $78882^{\circ}C$

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Example Question #1 : Calorimetry, Specific Heat, And Calculations

A 50g sample of a metal was heated to 95^oC then quickly transferred to an insulated container containing 50g of H_2O at 25^oC . The final temperature of the H_2O was 30^oC .

Which of the following can be concluded?

Possible Answers:

The specific heat of the metal is greater than that of the water

None of the other answers

The water gained an amount of thermal energy that was more than the amount of thermal energy lost by the metal

The metal lost an amount of thermal energy that was more than the amount of thermal energy gained by the water

The specific heat of the water is greater than that of the metal

Correct answer:

The specific heat of the water is greater than that of the metal

Explanation:

When the heated metal is placed in the container of the cooler water there will be a transfer of thermal energy from the metal to the water. This transfer will occur towards an equilibrium of thermal energy in the water and in the metal. Thus we can conclude that the amount of thermal energy lost by the metal will equal the amount of thermal energy gained by the water. However we notice that the water increases by only 5^oC and the metal decreases by 65^oC. This is becasue of the difference of the specific heats of these substances. The specific heat capacity of a substance is the heat required to increase the temperature of 1g of a substance by 1^oC. The metal can be conluded to have a smaller specific heat than the water because the same amount of energy transfer led to a much larger change in termperature for the metal as compared to the water.

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Example Question #3 : Calorimetry, Specific Heat, And Calculations

Which of the following is the correct molar specific heat of water used when making calculations involving a calorimeter?

Possible Answers:

4.184 J/gK

41.84 J/gK

0.4184 J/gK

418.4 J/gK

Correct answer:

4.184 J/gK

Explanation:

4.184 J/gK is the cited value for the specific heat of water and should be memorized. This is used during calorimeter calculations, specifically when using the equation q= mc delta(T).

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Example Question #3 : Thermodynamics

$\small C_{ice} = 0.5 \frac{cal}{g^{\circ}C}$

$\small C_{water} = 1.0\frac{cal}{g^{\circ}C}$

$\small \Delta H_{fusion} = 80\frac{cal}{g}$

How much energy is needed to raise the temperature of five grams of ice from $\small -10^oC$ to $\small 35^oC$ ?

Possible Answers:

$\small 120cal$

$\small 600 cal$

$\small 200cal$

$\small 280cal$

Correct answer:

$\small 600 cal$

Explanation:

This question involves the total energy needed for three different processes: the temperature raise from $\small -10^oC$ to $\small 0^oC$ , the melting of the ice, and the temperature raise from $\small 0^oC$ to $\small 35^oC$ . For the first and third transitions we will use the equation $\small q=mC\Delta T$ . For the melting of ice, we will use the equation $q=m\Delta H_{fusion}$ .

1. $\small \small q =m_{ice}C_{ice}\Delta T = (5g)(0.5\frac{cal}{g^{\circ}C})(10^{\circ}C) = 25 cal$

2. $\small q = m\Delta H_{fusion} = (5g)(80\frac{cal}{g}) = 400 cal$

3. $\small q = m_{water}C_{water}\Delta T = (5g)(1.0\frac{cal}{g^{\circ}C})(35^{\circ}C) = 175 cal$

Finally, we will need to sum the energy required for each step to find the total energy.

25cal+400cal+175cal=600cal

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Example Question #1 : Calorimetry, Specific Heat, And Calculations

You want to prepare a cup of tea. To do so, you pour $100\,mL$ of tap water at $20^{0}C$ in a cup that does not absorb microwave radiation and heat it in a microwave oven at $700\,W$ of power. If you assume a density of $1.00\, \frac{g}{mL}$ for the water and know that its specific heat capacity is $4.184\, \frac{J}{g^{o}C}$ , what time do you need to set in the microwave oven to heat the water to $60^{o}C$ ?

Possible Answers:

24s

700W of power is not enough to heat that a amount of water to the desired temperature

23.9s

40s

Correct answer:

23.9s

Explanation:

$Q=m\cdot C\cdot\Delta T$

Since the density of water is $1.00\, \frac{g}{mL}$ , the mass of $100\,mL$ is $100\,g$ . Plug in known values to the equation and solve.

$100g\cdot\frac{4.184J}{g^{\text o}C}\cdot \left ( 60 - 20 \right )^{\text o}C=16736J$

Use the formula below to find the time needed to heat up the sample of water in the microwave:

$P=\frac{\Delta W}{\Delta t}$

$\Delta t=\frac{16736J}{700W}=23.9s$

Our answer must contain three significant figures.

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Example Question #1 : Thermodynamics

How much heat is needed to raise grams of aluminum by $20^{o}\:C$ ?

$C_{aluminum}=0.900\: \frac{J}{g\cdot^{o}\textrm{C}}$ .

Possible Answers:

$45\:J$

$90\:J$

$9\:J$

$120\:J$

Correct answer:

$90\:J$

Explanation:

To find the amount of heat needed to change the temperature of a given material by a certain amount, we'll need to use the equation for specific heat. The specific heat capacity of a compound represents the amount of energy necessary to raise gram of that substance by $1\: ^{o}C$ .

$q=mC\Delta T$

$q=(5\: grams)(0.900\: \frac{J}{g\cdot ^{o}C})(20\: ^{o}C)=90\:J$

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Example Question #4 : Thermodynamics

A 20g sample of iron at a temperature of 120^oC is placed into a container of water. There are 300 milliliters of water in the container at a temperature of 30^oC .

$\small C_{iron} = 0.444\frac{J}{g^\circ C}$

$\small C_{water} = 4.184\frac{J}{g^\circ C}$

$\small \rho_{water} = 1\frac{g}{mL}$

What is the final temperature of the water?

Possible Answers:

$\small 114.0^{\circ}C$

$\small 36.0^{\circ}C$

$\small 30.63^{\circ}C$

$\small 34.87^{\circ}C$

32.71^oC

Correct answer:

$\small 30.63^{\circ}C$

Explanation:

There are two things to note before solving for the final temperature.

1. The density of water allows us to say that 300 milliliters of water is the same thing as 300 grams of water.

$300mL*\frac{1g}{1mL}=300g$

2. Since the heat from the iron is being transferred to the water, we can say that the heat transfer is equal between both compounds. Since the heat is conserved in the system, we can set the two equations equal to one another.

$q=mC\Delta T$

$q_i=q_w\rightarrow m_iC_i\Delta T=m_wC_w\Delta T$

$\small m_{i}C_{i}(T_{initial}-T_{final}) = m_{w}C_{w}(T_{final}-T_{initial})$

Notice how the change in temperature for iron has been flipped in order to avoid a negative number.

$\small (20g)(0.444\frac{J}{g^{\circ}C})(120^{\circ}C-T_{f}) = (300g)(4.184\frac{J}{g^{\circ}C})(T_{f}-30^{\circ}C)$

$\small T_{f} = 30.63^{\circ}C$

Because water has a much higher heat capacity compared to iron, the temperature of the water is not changed significantly.

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AP Chemistry : Thermodynamics

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Thermodynamics

Example Question #1 : Calorimetry, Specific Heat, And Calculations

Example Question #1 : Calorimetry, Specific Heat, And Calculations

Example Question #2 : Calorimetry, Specific Heat, And Calculations

Example Question #1 : Calorimetry, Specific Heat, And Calculations

Example Question #3 : Calorimetry, Specific Heat, And Calculations

Example Question #3 : Thermodynamics

Example Question #1 : Calorimetry, Specific Heat, And Calculations

Example Question #1 : Thermodynamics

Example Question #4 : Thermodynamics

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