Library Guides: Chemistry Textbook: Self-Assessments

1. Matter and MeasurementsToggle Dropdown
2. Atoms, Molecules and IonsToggle Dropdown
3. Composition of Substances and SolutionsToggle Dropdown
4. Stoichiometry of Chemical ReactionsToggle Dropdown
5. Thermochemistry
6. GasesToggle Dropdown
7. Chemical Bonding and Molecular GeometryToggle Dropdown
8. Liquids, Solids, and Modern MaterialsToggle Dropdown
9. Solutions and Colligative PropertiesToggle Dropdown
10. KineticsToggle Dropdown
11. Chemical Equilibria and ApplicationsToggle Dropdown
12. ThermodynamicsToggle Dropdown
13. ElectrochemistryToggle Dropdown
14. AppendicesToggle Dropdown

Self-Assessments

Energy Basics I
Energy Basics II
Energy Stoichiometry
Heat transfer (Calorimetry)
Reaction enthalpy from heats of formation

A piece of unknown substance weighs 44.7 g and requires 2.11 kJ to increase its temperature from 23.2 °C to 89.6 °C. What is the specific heat of the substance?

Choose 1 answer

0.63 J/g-^oC

-6.3 × 10³ J/g-^oC

6.3 × 10³ J/g-^oC

0.71 J/g-^oC

-0.71 J/g-^oC

Rearrange the equation for heat (q) to calculate the specific heat. Be sure to use appropriate units in the equation.

Specific heat cannot be a negative quantity!

Rearrange the equation for heat (q) to calculate the specific heat. Be sure to use appropriate units in the equation.

c = q / (m ΔT) = (2110 J) / ( 44.7g × (89.6-23.2)^oC ) = 0.71 J/(g ^oC)

Specific heat cannot be a negative quantity!

How much heat, in calories, must be added to a 75.0 g granite block with a specific heat of 0.790 J/g-K to increase its temperature from 25 °C to 80^oC?

Choose 1 answer

780

1600

3300

1.9 × 10⁴

1.4 × 10⁴

ΔT = 80^oC – 25^oC = 55^oC = 55 K. (Note that 1^oC ΔT = 1 K ΔT);
q = mcΔT = 75.0 g × (0.790 J/g K) × 55 K = 3259 J;
3259 J × (1 cal / 4.184 J) = 780 cal. Note that Joules in the numerator and denominator cancel.

Alternatively, you could use 0.790 J/gK = 0.790 J/g^oC

Heat gained or lost is calculated using q = mcΔT, where m is mass, c is specific heat and ΔT is final temperature minus the initial temperature. Use appropriate units in the equations.

Looks like you did not convert Joules to calories!

It seems you added 273 to ΔT in Celsius to get the difference in temperature in Kelvin. Note that 1^oC ΔT = 1 K ΔT. Therefore, ΔT = 55^oC = 55 K due to the reason shown below:
T_initial = 25^oC = 298 K, T_final = 80^oC = 353 K; ΔT = 353 – 298 = 55K

The conversion between Joules and calories is not used correctly. Use dimensional analysis set so that the unit in the denominator cancels the unit in the numerator.

An aluminum kettle weighs 1.05 kg. If the kettle loses 6540 J of heat to reach 42.1^oC, what is the kettle’s initial temperature? The specific heat of Aluminum is 0.89 J/g-K.

Choose 1 answer

49.1 ^oC

35.0 ^oC

47.6 ^oC

7040 ^oC

6956 ^oC

Great job!
specific heat, c = 0.89 J g^-1 K^-1; heat, q = -6540 J; mass, m = 1.05 kg = 1050 g;
ΔT = q / (m c) = (-6540 J)/ (1050 g ×89 J g^-1 K^-1) = -7.0 K = -7.0 ^oC;
ΔT = T_final - T_initial. Rearrange: T_initial = T_final - DT = 42.1^oC – (–7.0^oC) = 49.1^oC.

Either sign convention not followed in the q=mcΔT equation or error was made in calculating initial temperature after obtaining D
Note that heat (q) is negative as the heat is lost by the substance. Calculate DT firstThen use ΔT = T_final - T_initial.

Check calculation! A missing parenthesis may have caused calculation error!

Did not convert kg to grams!

Did not convert kg to grams. Also, the sign convention was not used appropriately.
Note that that heat (q) is negative as the heat is lost by the substance. Calculate ΔT first. Then use ΔT = T_final - T_initial.

The specific heat of copper is 0.385 J/g ^o What is it’s molar heat capacity?

Choose 1 answer

6.05 × 10^-3 J/mol^oC

225 J/mol^oC

165 J/mol^oC

24.5 J/mol^oC

11.2 J/mol^oC

Looks like you divided the specific heat by the molar mass of copper. Check if units cancel to give the desired unit!

Molar heat capacity is the heat needed per mole of substance per ^oC temperature change.

Looks like you divided molar mass of copper by the specific heat! Check if units cancel to give you the desired unit!

Molar heat capacity is the heat needed per mole of substance per ^oC temperature change.
From the definition of specific heat, to cause 1^oC change in temperature of copper, it takes 0.385 J per gram. We need to calculate how many Joules it will take per mole. A mole of Copper is 63.55 g. Therefore, for 63.55 g, it takes 63.55 times more heat.
385 J/(g^oC) × (63.55 g/mol)= 24.5 J/g^oC

Looks like you used the atomic number of copper for the calculation! Review how many grams is a mole of copper.

Calculate the heat capacity, in joules per degree, of 2.00 pound of aluminum metal. The specific heat of Aluminum is 0.89 J/g ^oC.

Choose 1 answer

1.02 × 10³

810

510

240

1.78

Use units to guide the calculation.

Heat capacity (J/^oC) is the amount of heat needed to raise change the temperature by 1 ^oC for a given amount of substance. From the definition of specific heat, 0.89 J is how much it takes per gram to change the temperature by 1^o So, for 2.00 lb (i.e. 908 g), it will take 0.89 J/g^oC × 908 g = 808 J/^oC = 810 J/^oC (after rounding). Notice that grams cancel and the desired unit, J^oC, is obtained in the calculation.

Looks like you have divided grams by the specific heat! Use units to guide your calculation. Units must cancel out to give the desired unit.

Looks you have calculated the molar specific heat! Molar mass is not needed for this problem!

You did not convert pounds to grams!

At a constant temperature, an ideal gas expands from 5.0 liters to 7.0 liters by a constant external pressure of 4.0 atm. Calculate the work involved in the process.
1 L atm = 101.33 J

Choose 1 answer

w_gas = 810 J

w_gas = −810 J

w_gas = 140 J

w_gas = −140 J

w_gas = 540 J

Check the sign. Formula for work is w = − P ΔV where ΔV = V_f − V_i.

Correct. Formula for work is w = − P ΔV
P = 4.0 atm, ΔV = V_f − V_i = +2 L. Substitute and convert L atm to Joules.
Notice that the calculation results in negative work which agrees with "work done by the system is negative" (the gas is the system. As the system expands from 5L to 7L, work is done by the system).

The system (gas) is expanding 5L to 7L. Work is done by the system, so, work should be negative!
Also, substitute correctly in the formula for work, w = − P ΔV, and solve.

Substitue correctly in the formula for work, w = − P ΔV

The system (gas) is expanding 5L to 7L. Work is done by the system, so, work should be negative!
Also, substitute correctly in the formula for work, w = − P ΔV, and solve.

What is the change in internal energy of a combustion engine that absorbs 100 J of heat while 130 J of work is done by the engine?

Choose 1 answer

+230 J

-230 J

+30 J

-30 J

0 J

The sign for work is not identified correctly!

The sign for heat is not identified correctly!

The signs for heat and work are not identified correctly!

Correct! It is important to identify q and w along with the sign.
q = +100 J (system absorbed heat) and w = −130 J (work is done by the system). Substitute in ΔE = q + w.

When system absorbs heat it's internal energy increases; when work is done by the system, system loses it's internal energy. The total change in internal energy is a combination change in internal energy due to heat and work. Use ΔE = q + w to solve.

A system releases 260 J of heat and at the same time 450 J of work is done on the system by the surroundings. What is the change in internal energy of the system?

Choose 1 answer

+710 J

-710 J

+190 J

-190 J

+130 J

The sign for heat is not identified correctly!

The sign for work is not identified correctly!

Correct! It is important to identify q and w along with the sign.
q = −260 J (system loses heat) and w = +450 J (work is done on the system). Substitute in ΔE = q + w.

The signs for heat and work are not identified correctly!

Use ΔE = q + w equation.

A system undergoes an increase in internal energy of 90 J and at the same time performs 55 J of work. Which of the following is true about the heat transferred during the process?

Choose 1 answer

the system absorbs 35 J of heat during the process.

the system releases 35 J of heat during the process.

the system absorbs 145 J of heat during the process.

the system releases 145 J of heat during the process.

not enough information to determine.

Sign for work is not determined correctly. Please read the statement about work carefully!

Sign for ΔE is not determined correctly. Note that internal energy has increased.

Correct! ΔE= +90 J (since internal energy increased), w = −55 J (since work is done by the system).
Solve for q using ΔE= q + w.
q = +145 J means system absorbed 145 J of heat during the process.

You must practice reading the statements and determining the signs correctly! Hint: Substitute in ΔE= q + w and solve for q.

Information about ΔE and work are provided, information about heat can be determined from the equation that relates ΔE, work and heat.

A gas enclosed in a metal container with a piston absorbs 1050 J of heat and 250 J of work is done by the gas while it expands at constant pressure. What is the change in internal energy and change in enthalpy of the gas during the process?

Choose 1 answer

ΔE = +800 J; ΔH = +1050 J

ΔE = +1050 J; ΔH = +800 J

ΔE = +800 J; ΔH = -250 J

ΔE = +1250 J; ΔH = +1050 J

ΔE = +1050 J; ΔH = +1250 J

ΔE is calculated using ΔE=q+w.
ΔH is calculated using ΔH = q_{at constant pressure}.

You may have switched ΔH and ΔE! ΔE = q+w, ΔH = q_{at constant pressure.}

ΔH is not correct! Recall that change in enthalpy (ΔH) is heat transferred at constant pressure.

ΔE is not correct! Recall that ΔE = q+w. Check signs for heat and work!

Both ΔE and ΔH values are not correct.
Hint: Internal energy change is energy change due to both heat and work.
Note that the heat is transferred at constant pressure in the problem. Enthalpy change is heat transferred at constant pressure.

Consider the reaction,
2Fe₂O₃(s) + 3C(s) → 4Fe(s) + 3CO₂(g) ΔH = 468.87 kJ.

How much heat is released at constant pressure when 35.0 g of iron is formed in the reaction?

Choose 1 answer

4100 kJ

1180 kJ

294 kJ

73.5 kJ

25.7 kJ

You probably did not convert grams to moles!

Hint: We are asked to find heat released at constant pressure (ie. ΔH) for forming 35.0 g of Fe.
Conversion factor from the equation and the associated ΔH is: 468.87 kJ = 4 mol Fe.

The coefficient for Fe is not used! Note that 468.87 kJ is the ΔH for forming 4 mol Fe!

We are asked to find heat released at constant pressure (ie. ΔH) for forming 35.0 g of Fe.
Conversion factor from the equation and the associated ΔH is: 468.87 kJ = 4 mol Fe.
Strategy: g of Fe --> moles of Fe --> kJ of heat.
$Solve using dimensional analysis steps. Start with 35.0 g Fe. Next fraction is 1 mol Fe over 55.85 g. Next fraction is 468.87 kJ over 4 mol Fe. Calcution yields 73.5 kJ$

Hint: We are asked to find heat released at constant pressure (ie. ΔH) for forming 35.0 g of Fe.
Conversion factor from the equation and the associated ΔH is: 468.87 kJ = 4 mol Fe.

Consider the reaction,
2Al(s) + Fe₂O₃(s) → 2Fe(s) + Al₂O₃(s) ΔH = -851.5 kJ.

If 18.2 g of powdered aluminum is allowed to react with excess Fe₂O₃, how much heat is produced at constant pressure?

Choose 1 answer

85.2 kJ

144 kJ

287 kJ

574 kJ

1150 kJ

Hint: We are asked to find heat released at constant pressure (ie. ΔH) for reacting 18.2 g of Al.
Conversion factor from the equation and the associated ΔH is: 2 mol Al = 851.5 kJ produced.

You possibly made an error in calculating the molar mass. Conversion factors: 1 mol Al = 26.98 g Al; 2 mol Al = 851.5 kJ produced.

We are asked to find heat released at constant pressure (ie. ΔH) for reacting 18.2 g of Al. Note: excess Fe₂O₃ simply suggests that 18.2 g Al reacted completely.
Conversion factor from the equation and the associated ΔH is: 2 mol Al = -851.5 kJ (note: the negative sign means heat is produced).
$Solve using dimensional analysis steps. Start with 18.2 g Al. Next fraction is 1 mol Al over 26.98 g. Next fraction is -851.5 kJ over 2 mol Al. Calcution yields -287 kJ, where the negative sign suggests that 287 kJ is produced.$

Coefficient for Aluminum is not used! Note that 851.5 kJ is released when 2 mol of Aluminum reacts.

We are asked to find heat released at constant pressure (ie. ΔH) for reacting 18.2 g of Al.
Conversion factor from the equation and the associated ΔH is: 2 mol Al = 851.5 kJ produced

Consider the reaction,
2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -484 kJ.

If 282 kJ of heat is released in the reaction at constant pressure, how many grams of water are formed during the process?

Choose 1 answer

21 g

10 g

5.1 g

1.1 g

0.57 g

Correct! The ΔH of the equation suggests that 484 kJ of heat is released when 2 moles of H₂O are formed. If only 282 kJ is released, we can calculate how many moles of H₂O are formed and then change it to grams. That is, kJ --> mol H₂O --> grams of H₂O.
$Solve using dimensional analysis steps. Start with 282 kJ released. Next fraction is 2 mol H2O over 484 kJ released. Next fraction is 18.02 g over 1 mol H2O. Calculation yields 21 g H2O.$

Please check your conversion factors! 2 mol H2O should appear in only one conversion factor!

Think: The ΔH of the equation suggests that 484 kJ of heat is released when 2 moles of H₂O are formed. If only 282 kJ is released, what can say about the amount of H₂O formed?

You probably did not convert moles of H₂O to grams.

Think: The ΔH of the equation suggests that 484 kJ of heat is released when 2 moles of H₂O are formed. If only 282 kJ is released, what can say about the amount of H₂O formed?

Choose 1 answer

-1180 kJ

-1134 kJ

822.2 kJ

776.3kJ

628.5 kJ

This is feedback for Option-A

This is feedback for Option-B

This is feedback for Option-C

This is feedback for Option-D

This is feedback for Option-E

<< Previous: End-of-Chapter Problems
Next: 6. Gases >>

Chemistry Textbook

Self-Assessments

A piece of unknown substance weighs 44.7 g and requires 2.11 kJ to increase its temperature from 23.2 °C to 89.6 °C. What is the specific heat of the substance?

How much heat, in calories, must be added to a 75.0 g granite block with a specific heat of 0.790 J/g-K to increase its temperature from 25 °C to 80oC?

An aluminum kettle weighs 1.05 kg. If the kettle loses 6540 J of heat to reach 42.1oC, what is the kettle’s initial temperature? The specific heat of Aluminum is 0.89 J/g-K.

The specific heat of copper is 0.385 J/g o What is it’s molar heat capacity?

Calculate the heat capacity, in joules per degree, of 2.00 pound of aluminum metal. The specific heat of Aluminum is 0.89 J/g oC.

At a constant temperature, an ideal gas expands from 5.0 liters to 7.0 liters by a constant external pressure of 4.0 atm. Calculate the work involved in the process. 1 L atm = 101.33 J

What is the change in internal energy of a combustion engine that absorbs 100 J of heat while 130 J of work is done by the engine?

A system releases 260 J of heat and at the same time 450 J of work is done on the system by the surroundings. What is the change in internal energy of the system?

A system undergoes an increase in internal energy of 90 J and at the same time performs 55 J of work. Which of the following is true about the heat transferred during the process?

A gas enclosed in a metal container with a piston absorbs 1050 J of heat and 250 J of work is done by the gas while it expands at constant pressure. What is the change in internal energy and change in enthalpy of the gas during the process?

Consider the reaction, 2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g) ΔH = 468.87 kJ.How much heat is released at constant pressure when 35.0 g of iron is formed in the reaction?

Consider the reaction, 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) ΔH = -851.5 kJ.If 18.2 g of powdered aluminum is allowed to react with excess Fe2O3, how much heat is produced at constant pressure?

Consider the reaction, 2H2(g) + O2(g) → 2H2O(g) ΔH = -484 kJ.If 282 kJ of heat is released in the reaction at constant pressure, how many grams of water are formed during the process?

For reaction, 2H2(g) + O2(g) → 2H2O(g), ΔH = -484 kJ.What is the enthalpy change associated with the following reaction?6H2(g) + 3O2(g) → 6H2O(g)

A 70.0-g piece of metal at 80.0 °C is placed in 100.0 mL of water at 22.0 °C contained in a coffee-cup calorimeter. The final temperature of water was found to be 24.6 °C. What is the specific heat of the metal? (Assume that no heat is transferred to the surroundings.)

A 151.0 g of coolant having a specific heat of 3.18 J/goC is mixed with 100.0 g of water with an initial temperature of 18.2 °C. If the final temperature of the water is 34.0 °C, what was the initial temperature of the coolant? (Assume that no heat is transferred to the surroundings.)

If a 30.0 g piece of copper pipe at 80.0 °C is placed in 100.0 g of water at 27.0 °C, what is the final temperature? Assume that no heat is transferred to the surroundings.Specific heat of copper is 0.385 J/go

Which of the following is Hess’s law?

Which of the following reactions represent the reaction for standard enthalpy of formation of bromomethane gas (CH3Br)?

Using the table of standard enthalpies of formation, calculate the standard enthalpy (ΔHoreaction) of the following reaction:4NH3(g)+5O2(g)→6H2O(g)+4NO(g)

How much heat, in calories, must be added to a 75.0 g granite block with a specific heat of 0.790 J/g-K to increase its temperature from 25 °C to 80^oC?

An aluminum kettle weighs 1.05 kg. If the kettle loses 6540 J of heat to reach 42.1^oC, what is the kettle’s initial temperature? The specific heat of Aluminum is 0.89 J/g-K.

The specific heat of copper is 0.385 J/g ^o What is it’s molar heat capacity?

Calculate the heat capacity, in joules per degree, of 2.00 pound of aluminum metal. The specific heat of Aluminum is 0.89 J/g ^oC.

At a constant temperature, an ideal gas expands from 5.0 liters to 7.0 liters by a constant external pressure of 4.0 atm. Calculate the work involved in the process.
1 L atm = 101.33 J

Consider the reaction,
2Fe₂O₃(s) + 3C(s) → 4Fe(s) + 3CO₂(g) ΔH = 468.87 kJ.

How much heat is released at constant pressure when 35.0 g of iron is formed in the reaction?

Consider the reaction,
2Al(s) + Fe₂O₃(s) → 2Fe(s) + Al₂O₃(s) ΔH = -851.5 kJ.

If 18.2 g of powdered aluminum is allowed to react with excess Fe₂O₃, how much heat is produced at constant pressure?

Consider the reaction,
2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -484 kJ.

If 282 kJ of heat is released in the reaction at constant pressure, how many grams of water are formed during the process?

For reaction, 2H₂(g) + O₂(g) → 2H₂O(g), ΔH = -484 kJ.

What is the enthalpy change associated with the following reaction?
6H₂(g) + 3O₂(g) → 6H₂O(g)

A 151.0 g of coolant having a specific heat of 3.18 J/g^oC is mixed with 100.0 g of water with an initial temperature of 18.2 °C. If the final temperature of the water is 34.0 °C, what was the initial temperature of the coolant? (Assume that no heat is transferred to the surroundings.)

If a 30.0 g piece of copper pipe at 80.0 °C is placed in 100.0 g of water at 27.0 °C, what is the final temperature? Assume that no heat is transferred to the surroundings.
Specific heat of copper is 0.385 J/g^o

Which of the following reactions represent the reaction for standard enthalpy of formation of bromomethane gas (CH₃Br)?

Using the table of standard enthalpies of formation, calculate the standard enthalpy (ΔH^o_reaction) of the following reaction:
4NH₃(g)+5O₂(g)→6H₂O(g)+4NO(g)