Internal Energy and Thermodynamics Problem Set Analysis

Thermodynamics of Vented Gas Systems and Ideal Gas Behavior

• Problem Context (Problem 3.3):     * The scenario involves an empty oven that is being heated from an initial state to a final state while being vented to the atmosphere.     * System Volume ($V$): The volume of the oven is constant at $0.150\,m^3$.     * Venting Mechanism: The oven is specifically vented so that the internal air pressure ($P_{internal}$) always remains equal to the environmental air pressure ($P_{environment}$).

• Initial Conditions:     * Initial Temperature ($T_i$): $296\,K$.     * Initial Atmospheric Pressure ($P_i$): $1.00 \times 10^5\,Pa$.

• Final Conditions:     * Final Temperature ($T_f$): $453\,K$.     * Final Atmospheric Pressure ($P_f$): $9.50 \times 10^4\,Pa$.     * Pressure Change Note: The decrease in atmospheric pressure is attributed to changing weather conditions occurring during the heating process.

• Calculation Objective: Moles of Air Escaping:     * The goal is to calculate the specific quantity of air (in moles) that leaves the oven during the transition from the initial to the final state.     * Mathematical Approach: Use the Ideal Gas Law equation $PV = nRT$.     * Calculate initial moles ($n_i$) using $n_i = \frac{P_i V}{R T_i}$.     * Calculate final moles ($n_f$) using $n_f = \frac{P_f V}{R T_f}$.     * The amount of air leaving is determined by the difference: $\Delta n = n_i - n_f$.

Energy Transformation and Internal Energy Generation

• Kinetic to Internal Energy Conversion (Problem 4.1):     * This problem analyzes the conversion of mechanical energy into internal energy (heat) when an object is brought to a sudden stop.     * Object Specifications: A lead bullet with a mass ($m$) of $20\,g$ ($0.02\,kg$).     * Initial State: The bullet is traveling horizontally at a velocity ($v$) of $7 \times 10\,m/s$.     * Final State: The bullet comes to a complete stop ($v_f = 0\,m/s$) upon striking a metal plate.

• Calculation Methodology:     * The amount of internal energy generated is equal to the total kinetic energy lost by the bullet during the impact.     * Equation for Kinetic Energy ($K$): $K = \frac{1}{2} m v^2$.     * Substituting values: $\Delta U = \frac{1}{2} \times (0.02\,kg) \times (7 \times 10\,m/s)^2$.

Calorimetry and Thermal Equilibrium with Gravitational Potential Energy

• Scenario Overview (Problem 4.2):     * A high-mass steel ball is dropped into a container of water. The final temperature of the system must be calculated after thermal equilibrium is reached.

• System Components and Initial Properties:     * Steel Ball:         * Mass ($m_s$): $7.3\,kg$.         * Initial Temperature ($T_{si}$): $15.2\,^{\circ}C$.         * Initial Height ($h$): $10\,m$.     * Water (in an insulated container):         * Volume ($V_w$): $4.5\,L$.         * Initial Temperature ($T_{wi}$): $10.1\,^{\circ}C$.

• Physical Constants and Assumptions:     * Density of Water ($\rho$): $1\,kg/L$.     * Specific Heat Capacity of Water ($c_w$): $4186\,J/(kg \cdot K)$.     * Specific Heat Capacity of Steel ($c_s$): $450\,J/(kg \cdot K)$.     * Mass of Water Calculation: Since $\rho = 1\,kg/L$, the mass of water ($m_w$) is $4.5\,kg$.     * Energy Conservation Assumption: It is assumed that no energy is lost to the container itself and no water is lost through splashing.

• Energy Balance Principles:     * The system reaches equilibrium at a final temperature ($T_f$).     * The total energy contributed to the system includes the initial thermal energy of both substances and the mechanical potential energy ($PE$) of the ball being converted into thermal energy upon impact.     * Potential Energy Equation: $PE = m_s g h$, where $g$ is the acceleration due to gravity ($9.8\,m/s^2$).     * Heat Transfer Equation ($Q$): $Q = m c \Delta T$.     * Conservation of Energy Equation: $m_w c_w (T_f - T_{wi}) + m_s c_s (T_f - T_{si}) = m_s g h$.     * This equation accounts for the heat absorbed by the water and the heat change of the ball, balanced against the converted gravitational potential energy.

Summary of Provided Physical Data and Constraints

• Document Context: These problems are part of the PHY131 Semester Test 2, dated 2025.
• Oven Parameters (Problem 3.3):     * Dimension: $0.150\,m^3$.     * Thermal Range: $296\,K$ to $453\,K$.     * Pressure Range: $1.00 \times 10^5\,Pa$ to $9.50 \times 10^4\,Pa$.
• Ballistics Parameters (Problem 4.1):     * Lead Mass: $20\,g$.     * Velocity: $7 \times 10\,m/s$.
• Equilibrium Parameters (Problem 4.2):     * Steel Mass: $7.3\,kg$.     * Water Volume: $4.5\,L$.     * Height: $10\,m$.     * Water Sp. Heat: $4186\,J/(kg \cdot K)$.     * Steel Sp. Heat: $450\,J/(kg \cdot K)$.