Ap Physics 2 Thermodynamics Practice Problems

Thermodynamics, the study of energy transfer and its transformations, is a cornerstone of physics. For AP Physics 2 students, mastering thermodynamics concepts and problem-solving techniques is crucial for exam success. This article provides a comprehensive set of practice problems, covering key topics from the laws of thermodynamics to heat engines and entropy. These problems are designed to challenge your understanding and build your confidence in tackling any thermodynamics question.

Fundamental Concepts in Thermodynamics

Before diving into the problems, let's revisit some essential principles:

• The First Law of Thermodynamics: This law states that the change in internal energy (ΔU) of a system equals the heat added to the system (Q) minus the work done by the system (W): ΔU = Q - W. It's essentially the law of conservation of energy applied to thermodynamic systems.
• The Second Law of Thermodynamics: This law introduces the concept of entropy (S), a measure of disorder in a system. It states that the entropy of an isolated system always increases or remains constant in a reversible process: ΔS ≥ 0. This law implies that heat cannot spontaneously flow from a colder body to a hotter body without external work being done.
• The Third Law of Thermodynamics: This law states that as the temperature of a system approaches absolute zero (0 Kelvin), all processes cease, and the entropy of the system approaches a minimum or zero value.
• Thermodynamic Processes:
  • Isothermal Process: A process occurring at constant temperature (ΔT = 0).
  • Adiabatic Process: A process occurring without heat exchange with the surroundings (Q = 0).
  • Isobaric Process: A process occurring at constant pressure (ΔP = 0).
  • Isochoric (or Isovolumetric) Process: A process occurring at constant volume (ΔV = 0).
• Heat Engines and Refrigerators: Heat engines convert thermal energy into mechanical work, while refrigerators transfer heat from a cold reservoir to a hot reservoir. Their efficiency is a critical parameter.
• Heat Transfer: Heat can be transferred through conduction, convection, and radiation.

Practice Problems: Laws of Thermodynamics

Let's begin with problems based on the fundamental laws of thermodynamics:

Problem 1:

A gas is enclosed in a cylinder fitted with a movable piston. The initial volume of the gas is 0.1 m³, and the pressure is 2.0 x 10⁵ Pa. If 1.5 x 10⁴ J of heat is added to the gas, and the volume increases to 0.15 m³ while the pressure remains constant, calculate:

a) The work done by the gas.

b) The change in internal energy of the gas.

Solution:

a) Since the pressure is constant (isobaric process), the work done is:

W = PΔV = (2.0 x 10⁵ Pa) * (0.15 m³ - 0.1 m³) = 1.0 x 10⁴ J

b) Using the first law of thermodynamics:

ΔU = Q - W = (1.5 x 10⁴ J) - (1.0 x 10⁴ J) = 5.0 x 10³ J

Problem 2:

A 50.0 g piece of metal at 85.0°C is placed in 100.0 g of water at 22.0°C. The final temperature of the water and metal is 25.6°C. Assuming no heat is lost to the surroundings, determine the specific heat capacity of the metal. (The specific heat capacity of water is 4186 J/kg·°C).

Solution:

Heat lost by the metal = Heat gained by the water

• Q_metal = m_metal * c_metal * ΔT_metal = (0.050 kg) * c_metal * (25.6°C - 85.0°C)
• Q_water = m_water * c_water * ΔT_water = (0.100 kg) * (4186 J/kg·°C) * (25.6°C - 22.0°C)

Equating the magnitudes:

(0.050 kg) * c_metal * (59.4°C) = (0.100 kg) * (4186 J/kg·°C) * (3.6°C)

c_metal = (0.100 kg * 4186 J/kg·°C * 3.6°C) / (0.050 kg * 59.4°C) ≈ 507 J/kg·°C

Problem 3:

A 2.0 mol sample of an ideal gas expands isothermally from a volume of 10.0 L to 25.0 L at a temperature of 300 K. Calculate:

a) The work done by the gas during the expansion.

b) The heat absorbed by the gas during the expansion.

Solution:

a) For an isothermal process, the work done is:

W = nRT * ln(V₂/V₁) = (2.0 mol) * (8.314 J/mol·K) * (300 K) * ln(25.0 L / 10.0 L)

W ≈ 4570 J

b) Since the process is isothermal, ΔT = 0, and therefore ΔU = 0. Using the first law:

Q = W + ΔU = 4570 J + 0 = 4570 J

Problem 4:

An ideal gas undergoes an adiabatic compression from a volume of 5.0 L to 1.0 L. The initial pressure is 1.0 atm, and the adiabatic index (γ) for the gas is 1.4. Calculate the final pressure.

Solution:

For an adiabatic process, P₁V₁^γ = P₂V₂^γ

P₂ = P₁ * (V₁/V₂)^γ = (1.0 atm) * (5.0 L / 1.0 L)^1.4 ≈ 9.5 atm

Problem 5:

A 0.5 kg block of ice at -10°C is heated until it completely melts and turns into water at 20°C. Calculate the total amount of heat required. (Specific heat capacity of ice = 2100 J/kg·°C, Latent heat of fusion of ice = 3.34 x 10⁵ J/kg, Specific heat capacity of water = 4186 J/kg·°C).

Solution:

The process involves three stages:

1. Heating the ice from -10°C to 0°C:

   Q₁ = m * c_ice * ΔT = (0.5 kg) * (2100 J/kg·°C) * (10°C) = 10500 J

2. Melting the ice at 0°C:

   Q₂ = m * L_f = (0.5 kg) * (3.34 x 10⁵ J/kg) = 167000 J

3. Heating the water from 0°C to 20°C:

   Q₃ = m * c_water * ΔT = (0.5 kg) * (4186 J/kg·°C) * (20°C) = 41860 J

Total heat required:

Q_total = Q₁ + Q₂ + Q₃ = 10500 J + 167000 J + 41860 J = 219360 J

Practice Problems: Heat Engines and Refrigerators

Now, let's explore problems related to heat engines and refrigerators:

Problem 6:

A heat engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. In each cycle, it absorbs 1200 J of heat from the hot reservoir.

a) What is the maximum possible efficiency of this engine?

b) How much work does the engine perform in each cycle if it operates at its maximum efficiency?

c) How much heat is exhausted to the cold reservoir in each cycle?

Solution:

a) The maximum efficiency (Carnot efficiency) is:

η<sub>max</sub> = 1 - (T<sub>cold</sub> / T<sub>hot</sub>) = 1 - (300 K / 500 K) = 0.4 or 40%

b) Work done at maximum efficiency:

W = η<sub>max</sub> * Q<sub>hot</sub> = 0.4 * 1200 J = 480 J

c) Heat exhausted to the cold reservoir:

Q<sub>cold</sub> = Q<sub>hot</sub> - W = 1200 J - 480 J = 720 J

Problem 7:

A refrigerator has a coefficient of performance (COP) of 5.0. If the refrigerator removes 300 J of heat from the cold reservoir in each cycle, how much work is required to operate the refrigerator? How much heat is exhausted to the hot reservoir?

Solution:

COP = Q_cold / W

W = Q_cold / COP = 300 J / 5.0 = 60 J

Heat exhausted to the hot reservoir:

Q_hot = Q_cold + W = 300 J + 60 J = 360 J

Problem 8:

A Carnot engine operates between temperatures of 600 K and 300 K. During each cycle, it absorbs 1000 J of heat from the hot reservoir. Calculate:

a) The work done by the engine per cycle.

b) The heat rejected to the cold reservoir per cycle.

c) The change in entropy of the hot reservoir per cycle.

d) The change in entropy of the cold reservoir per cycle.

Solution:

a) Efficiency of the Carnot engine:

η = 1 - (T<sub>cold</sub> / T<sub>hot</sub>) = 1 - (300 K / 600 K) = 0.5

Work done per cycle:

W = η * Q<sub>hot</sub> = 0.5 * 1000 J = 500 J

b) Heat rejected to the cold reservoir:

Q<sub>cold</sub> = Q<sub>hot</sub> - W = 1000 J - 500 J = 500 J

c) Change in entropy of the hot reservoir (heat is leaving, so negative):

ΔS<sub>hot</sub> = -Q<sub>hot</sub> / T<sub>hot</sub> = -1000 J / 600 K ≈ -1.67 J/K

d) Change in entropy of the cold reservoir (heat is entering, so positive):

ΔS<sub>cold</sub> = Q<sub>cold</sub> / T<sub>cold</sub> = 500 J / 300 K ≈ 1.67 J/K

Note: The total change in entropy for a reversible Carnot cycle is zero: ΔS_hot + ΔS_cold = 0.

Practice Problems: Entropy

Let's explore some problems focusing on entropy calculations:

Problem 9:

A 1.0 kg block of aluminum at 100°C is placed in a large lake at 15°C. Assuming the lake's temperature remains constant, calculate the change in entropy of the aluminum and the change in entropy of the lake. (Specific heat capacity of aluminum = 900 J/kg·°C).

Solution:

First, find the heat lost by the aluminum:

Q = m * c * ΔT = (1.0 kg) * (900 J/kg·°C) * (15°C - 100°C) = -76500 J (negative because heat is lost)

Change in entropy of the aluminum:

ΔS_aluminum ≈ Q / T_avg = -76500 J / [(100 + 15 + 273.15 + 273.15)/2 K] ≈ -76500 J / 330.65 K ≈ -231.3 J/K

(Note: A more accurate calculation would involve integrating dQ/T, but this approximation is often sufficient).

Change in entropy of the lake:

The lake gains the same amount of heat (76500 J) that the aluminum loses. Since the lake is large, its temperature remains constant.

ΔS_lake = Q / T = 76500 J / (15 + 273.15) K = 76500 J / 288.15 K ≈ 265.5 J/K

Notice that the total entropy change (ΔS_aluminum + ΔS_lake) is positive, as required by the second law of thermodynamics for irreversible processes.

Problem 10:

Two moles of an ideal gas undergo a free expansion from an initial volume of 10.0 L to a final volume of 25.0 L. Calculate the change in entropy of the gas.

Solution:

For a free expansion (also called a Joule expansion), no work is done (W = 0) and no heat is exchanged (Q = 0). Therefore, the internal energy remains constant (ΔU = 0), and the temperature remains constant (isothermal).

The change in entropy for an isothermal process is:

ΔS = nR * ln(V₂/V₁) = (2.0 mol) * (8.314 J/mol·K) * ln(25.0 L / 10.0 L) ≈ 15.3 J/K

Problem 11:

A 100 g piece of copper at 80°C is placed in 200 g of water at 20°C. The system is thermally isolated. Calculate the change in entropy of the copper, the water, and the total change in entropy of the system when they reach thermal equilibrium. (Specific heat capacity of copper = 385 J/kg·°C, Specific heat capacity of water = 4186 J/kg·°C).

Solution:

First, find the final equilibrium temperature (T_f):

Heat lost by copper = Heat gained by water

(0.100 kg) * (385 J/kg·°C) * (80°C - T_f) = (0.200 kg) * (4186 J/kg·°C) * (T_f - 20°C)

3080 - 38.5T_f = 837.2T_f - 16744

875.7T_f = 19824

T_f ≈ 22.64°C

Now, calculate the change in entropy for each:

• Copper:

  Q_copper = (0.100 kg) * (385 J/kg·°C) * (22.64°C - 80°C) = -2210.9 J

  ΔS_copper ≈ Q / T_avg = -2210.9 J / [(80 + 22.64 + 273.15 + 273.15)/2 K] ≈ -2210.9 J / 324.47 K ≈ -6.82 J/K

• Water:

  Q_water = (0.200 kg) * (4186 J/kg·°C) * (22.64°C - 20°C) = 2210.9 J

  ΔS_water ≈ Q / T_avg = 2210.9 J / [(20 + 22.64 + 273.15 + 273.15)/2 K] ≈ 2210.9 J / 284.47 K ≈ 7.77 J/K

Total change in entropy:

ΔS_total = ΔS_copper + ΔS_water ≈ -6.82 J/K + 7.77 J/K ≈ 0.95 J/K

The total entropy change is positive, as expected for an irreversible process.

Tips for Solving Thermodynamics Problems

• Understand the Concepts: Thermodynamics relies on a few fundamental laws and definitions. Make sure you thoroughly understand these principles before attempting problems.
• Identify the Process: Determine what type of thermodynamic process is occurring (isothermal, adiabatic, isobaric, isochoric). This will guide your choice of equations.
• Draw a Diagram: Sketching a PV diagram can be helpful for visualizing the process and identifying the work done.
• Pay Attention to Units: Ensure all quantities are expressed in consistent units (e.g., Joules for energy, Kelvin for temperature, Pascals for pressure, cubic meters for volume).
• Use the Correct Sign Conventions: Be mindful of the sign conventions for heat (positive when added to the system) and work (positive when done by the system).
• Apply the First Law of Thermodynamics: This law is the foundation for many thermodynamics problems.
• Consider Entropy: Remember that the entropy of an isolated system never decreases.
• Check Your Answers: Make sure your answers are physically reasonable. For example, the efficiency of a heat engine cannot exceed the Carnot efficiency.
• Practice Regularly: The more problems you solve, the better you will become at recognizing patterns and applying the appropriate techniques.

Advanced Problems

Here are some more challenging problems to test your mastery of thermodynamics:

Problem 12:

A Carnot engine uses 1.0 mol of an ideal monatomic gas as its working substance. The cycle starts at point A with a pressure of 1.0 atm and a temperature of 300 K. The engine undergoes the following processes:

• A to B: Isothermal expansion to twice the initial volume.
• B to C: Adiabatic expansion until the temperature reaches 200 K.
• C to D: Isothermal compression.
• D to A: Adiabatic compression back to the initial state.

Calculate:

a) The pressure, volume, and temperature at each of the points A, B, C, and D.

b) The heat exchanged during the isothermal processes.

c) The work done during each process.

d) The efficiency of the engine.

Problem 13:

A well-insulated container is divided into two compartments by a partition. One compartment contains 2.0 mol of an ideal gas at a temperature of 300 K and a volume of 20.0 L. The other compartment contains 3.0 mol of the same ideal gas at a temperature of 400 K and a volume of 30.0 L. The partition is removed, and the gases are allowed to mix. Calculate:

a) The final temperature of the mixture.

b) The final volume of the mixture.

c) The change in entropy of the system.

Problem 14:

A heat pump is used to heat a house. The outside temperature is -5°C, and the inside temperature is 22°C. The heat pump delivers heat to the house at a rate of 7.0 kW. Assuming the heat pump operates at the Carnot limit, calculate:

a) The coefficient of performance of the heat pump.

b) The electrical power required to operate the heat pump.

Problem 15:

A reversible heat engine operates between two reservoirs at temperatures T₁ and T₂ (where T₁ > T₂). The working substance is an ideal gas. Derive an expression for the efficiency of the engine in terms of T₁ and T₂ for the following cycle:

• Isothermal expansion at T₁ from volume V₁ to volume V₂.
• Isochoric (constant volume) process from T₁ to T₂.
• Isothermal compression at T₂ from volume V₂ to volume V₁.
• Isochoric process from T₂ to T₁.

Conclusion

Mastering thermodynamics requires a solid understanding of the fundamental laws, a familiarity with different thermodynamic processes, and plenty of practice solving problems. These practice problems, ranging from basic to advanced, are designed to help you develop your problem-solving skills and prepare for the AP Physics 2 exam. Remember to focus on understanding the underlying concepts and applying them correctly. By diligently working through these problems and reviewing the key principles, you can confidently tackle any thermodynamics challenge. Good luck!