Chapter 06 Review Problems

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Thermochemistry Basics

A chemist is studying the process of melting sugar in a pot on a stove. If the chemist defines the sugar as the system, which of the following is best classified as the immediate surroundings?

  1. The heat flowing into the sugar
  2. The pot and the stove burner
  3. The air in the kitchen
  4. The temperature of the sugar
Solution

Answer: B

  • (a) Incorrect. Heat is not a physical entity. It is the transfer of thermal energy between the system and surroundings, so it cannot be classified as the surroundings.
  • (b) Correct. The immediate surroundings are whatever is in direct physical contact with the system. The sugar (system) sits directly in the pot, which is heated by the stove burner.
  • (c) Incorrect. The air in the kitchen is part of the surroundings, but it is not in immediate contact with the sugar. The pot is.
  • (d) Incorrect. Temperature is a property of the system, not a physical entity that can be classified as surroundings.

Concept: thermochemistry basics

Which of the following is not a type of potential energy (select all that apply)?

  1. Energy held in chemical bonds
  2. Energy resulting from intermolecular attractions
  3. Energy from the random motion of molecules
  4. Energy of a ball in motion
Solution

Answer: C, D

  • (a) Potential energy. Chemical bonds store energy due to the positions and interactions of atoms. Breaking bonds requires energy input; forming bonds releases energy.
  • (b) Potential energy. Intermolecular attractions (hydrogen bonding, dipole-dipole forces, London dispersion forces) represent stored energy based on molecular positions and interactions.
  • (c) Not potential energy. Random molecular motion is kinetic energy (thermal energy). This energy depends on motion, not position.
  • (d) Not potential energy. A ball in motion has kinetic energy. Potential energy would be a ball at rest at a height.

Concept: thermochemistry basics

Which of the following is true of heat (select all that apply)?

  1. Heat is a form of thermal energy.
  2. Heat is the transfer of thermal energy.
  3. Heat is the action of forces through a distance.
  4. A negative heat in the system means the surroundings loses energy.
  5. A negative heat in the system means the system loses energy.
Solution

Answer: B, E

  • (a) False. Heat is not a form of thermal energy. Thermal energy is what a system possesses; heat is the process of transferring thermal energy between system and surroundings.
  • (b) True. Heat is defined as the transfer of thermal energy due to a temperature difference.
  • (c) False. This is the definition of work, not heat.
  • (d) False. If qsys < 0, the system loses energy, so the surroundings gains energy.
  • (e) True. A negative q for the system means energy flows out of the system.

Concept: thermochemistry basics

Which of the following are false (select all that apply)?

  1. Energy can be converted from one type to another.
  2. Energy is the capacity to do work.
  3. Kinetic energy is energy resulting from condition, position, or composition.
  4. Potential energy is energy transferred between a system and its surroundings as a result of a temperature difference.
Solution

Answer: C, D

  • (a) True. Energy can be converted from one form to another (e.g., chemical to thermal, kinetic to potential). This is a fundamental principle of the First Law of Thermodynamics.
  • (b) True. This is a standard definition of energy.
  • (c) False. This describes potential energy, not kinetic energy. Kinetic energy is energy due to motion.
  • (d) False. This describes heat, not potential energy. Potential energy is stored energy due to position, condition, or composition.

Concept: thermochemistry basics

A block of ice absorbs heat and melts. The value q for the system is:

  1. Positive
  2. Negative
  3. Zero
  4. There is not enough information to determine.
Solution

Answer: A

  • (a) Correct. The ice absorbs heat from the surroundings. When a system gains energy, q > 0.
  • (b) Incorrect. A negative q would mean the system releases heat, but the problem states the ice absorbs heat.
  • (c) Incorrect. q = 0 would mean no heat transfer, but the ice is absorbing heat to melt.
  • (d) Incorrect. The problem provides enough information: the ice absorbs heat, so q must be positive.

Concept: thermochemistry basics

Thermochemistry, Heat

Suppose 1.30 kJ of energy is removed by a refrigeration unit from a 2.19 kg sample of brass (cp(brass) = 0.380 J g−1 °C) at a temperature of 24.6 °C. A scientist wants to calculate the new temperature of the sample.

  1. What are the surroundings in the above problem?
    1. the refrigeration unit
    2. the sample of brass
    3. the sample of brass and the refrigeration unit
    4. none of the above
  2. Which equation(s) applies/apply?
    1. qcal = –qrxn
    2. q = mcpΔT
    3. q = CcalΔT
    4. none of the above
  3. Will the change in energy of the system, qsys, be positive or negative?
    1. positive
    2. negative
  4. Which of the given values must be converted to match the units of the specific heat?
    1. energy ( q )
    2. mass ( m )
    3. temperature ( T )
  5. Fill in the correct values for the brass sample (system): energy (in J), mass (in g), and temperature (in °C).
    1. Energy:
    2. Mass:
    3. Temperature:
  6. What is the new temperature (in °C) of the sample of brass?
Solution

Answer:

  1. 1
  2. 2
  3. 2
  4. 1, 2
  5.  (1) −1.30 × 103 J (negative because the system loses energy); (2) 2.19 × 103 g; (3) 24.6 °C
  6. 23.0 °C (see work below)

\[ \begin{align*} q &= m(\mathrm{brass}) ~ c_p(\mathrm{brass}) ~ \Delta T(\mathrm{brass}) \\[0.5ex] &= m(\mathrm{brass}) ~ c_p(\mathrm{brass}) ~ \bigr[ T_{\mathrm{f}} - T_{\mathrm{i}} \bigr]~ \longrightarrow \\[0.5ex] T_{\mathrm{f}} &= \biggl\{ q ~ \bigr[ m(\mathrm{brass}) ~ c_p(\mathrm{brass}) \bigr]^{-1} \biggl\} ~+~ T_{\mathrm{i}}\\[0.5ex] &= \biggl\{ -1.30 \times 10^{3}~\mathrm{J}~ \biggr[ 2.19 \times 10^{3}~\mathrm{g} \left ( \dfrac{0.380~\mathrm{J}}{\mathrm{g~^{\circ}C}} \right ) \biggr]^{-1} \biggl\} ~+~ 24.6~^{\circ}\mathrm{C} \\[0.5ex] &= -1.\bar{5}62~^{\circ}\mathrm{C} + 24.6~^{\circ}\mathrm{C} \\[0.5ex] &= 23.\bar{0}37~^{\circ}\mathrm{C} = 23.0~^{\circ}\mathrm{C} \end{align*} \]

Concept: thermochemistry basics; heat; specific heat capacity

If a system cools from 299 K to 294 K, what is the sign of q?

  1. positive
  2. negative
  3. impossible to know
Solution

Answer: B

  • (a) Incorrect. A positive q means the system absorbs heat and its temperature increases. The system is cooling, not heating.
  • (b) Correct. When a system cools, it releases heat to the surroundings. Energy leaving the system means q < 0.
  • (c) Incorrect. The problem provides enough information. The direction of temperature change determines the sign of q.

Concept: thermochemistry basics; heat

A container holds 2.00 L of water. How much heat (in J in normalized scientific notation) is required to warm the water from 25.0 °C to 100.0 °C? (d(H2O) = 1.00 g mL−1; cp(H2O) = 4.184 J g−1 °C−1)

Solution

Answer: 6.28 × 105 J

Concept: heat capacity; temperature change

\[ \begin{align*} q &= mc_{\mathrm{p}}\Delta T \\[1.5ex] &= V\rho~c_{\mathrm{p}} \left (T_{\mathrm{f}} - T_{\mathrm{i}} \right ) \\[1.5ex] &= (2.0\bar{0}~\mathrm{L}) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \left ( \dfrac{1.0\bar{0}~\mathrm{g}}{\mathrm{mL}} \right ) \left ( \dfrac{4.18\bar{4}~\mathrm{J}}{\mathrm{g~^{\circ}C}} \right ) \left ( 100.\bar{0}~^{\circ}\mathrm{C} - 25.\bar{0}~^{\circ}\mathrm{C} \right ) \\[1.5ex] &= (2.0\bar{0}~\mathrm{L}) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \left ( \dfrac{1.0\bar{0}~\mathrm{g}}{\mathrm{mL}} \right ) \left ( \dfrac{4.18\bar{4}~\mathrm{J}}{\mathrm{g~^{\circ}C}} \right ) \left ( 75.\bar{0}~^{\circ}\mathrm{C} \right ) \\[1.5ex] &= 6.2\bar{7}60\times 10^{5}~\mathrm{J} \\[1.5ex] &= 6.28\times 10^{5}~\mathrm{J} \end{align*} \]

A 3.10 g piece of solid copper is initially −12.0 °C and absorbs 58.4 J of energy. What is the final temperature (in °C) of the copper? (Cp(Cu) = 24.4 J mol−1 °C−1)

Solution

Answer: 37.1 °C

Concept: heat capacity; temperature change

\[ \begin{align*} q &= mc_p\Delta T \\[1.5ex] &= m~C_p~M^{-1} \left ( T_{\mathrm{f}} - T_{\mathrm{i}} \right ) \longrightarrow \\[1.5ex] T_{\mathrm{f}} &= \biggl [ q m^{-1} C_{\mathrm{p}}^{-1} M \biggr ] + T_{\mathrm{i}} \\[1.5ex] &= \biggl [ \left ( 58.\bar{4}~\mathrm{J} \right ) \left ( \dfrac{1}{3.1\bar{0}~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{mol~^{\circ}C}}{24.\bar{4}~\mathrm{J}} \right ) \left ( \dfrac{63.5\bar{5}~\mathrm{g}}{\mathrm{mol}} \right ) \biggl ] ~+~ \left ( -12.\bar{0}~^{\circ}\mathrm{C} \right ) \\[1.5ex] &= 49.\bar{0}65~^{\circ}\mathrm{C} + \left ( -12.\bar{0}~^{\circ}\mathrm{C} \right ) \\[1.5ex] &= 37.\bar{0}65~^{\circ}\mathrm{C} \\[1.5ex] &= 37.1~^{\circ}\mathrm{C} \end{align*} \]

A container contains 135.5 g H2O(l) at 17.2 °C. A 32.2 g piece of solid Al at 47.1 °C is placed in the container. What is the final temperature (in °C) of the system? (cp(Al) = 0.897 J g−1 °C−1; cp(H2O) = 4.184 J g−1 °C−1)

Solution

Answer:

Concept: heat capacity; equilibrium temperature

\[ \begin{align*} q_{\mathrm{Al}} &= q_{\mathrm{H_2O}} \\[1.5ex] &= \\[1.5ex] &= \end{align*} \]

Solution

Answer:

Concept:

$$ \[\begin{align*} &= \\[1.5ex] &= \end{align*}\] $$