Thermodynamics and Heat Transfer: Specific Heat

Definition and Fundamentals of Specific Heat

  • Conceptual Definition: Specific heat is defined as the energy required to raise the temperature of a unit mass of a substance by exactly one degree.

  • The Thermodynamic "Price" Metaphor: Specific heat can be viewed as an "exchange rate." It describes how many kilojoules (kJkJ) must be "spent" to "buy" a temperature change of one degree for one kilogram (kgkg) of a specific material.

  • Unit Mass and Temperature Parameters:     * Mass (mm) = 1kg1\,kg     * Temperature change (ΔT\Delta T) = 1C1\,^{\circ}C

  • Common Units:     * kJ/kgCkJ/kg \cdot ^{\circ}C     * kJ/kgKkJ/kg \cdot K     * Note: Since the intervals for Celsius and Kelvin are identical (ΔC=ΔK\Delta ^{\circ}C = \Delta K), these units are interchangeable in the context of specific heat.

Relating Specific Heat to Energy States: Cv vs Cp

Thermodynamics distinguishes between two types of specific heats based on the physical constraints applied during heating.

  • Specific Heat at Constant Volume (cvc_v):     * Physical Constraint: Volume remains constant (V=CV = C).     * Energy Focus: Related to changes in internal energy (dudu).     * Mathematical Definition: cv=(uT)vc_v = \left( \frac{\partial u}{\partial T} \right)_v. This represents the change in internal energy with respect to temperature at a constant volume.     * Physical Device Analogy: A rigid tank.

  • Specific Heat at Constant Pressure (cpc_p):     * Physical Constraint: Pressure remains constant (P=CP = C).     * Energy Focus: Related to changes in enthalpy (dhdh).     * Mathematical Definition: cp=(hT)pc_p = \left( \frac{\partial h}{\partial T} \right)_p. This represents the change in enthalpy with respect to temperature at a constant pressure.     * Physical Device Analogy: A freely moving piston.

  • The "Mechanical Tax" (Why c_p > c_v):     * When heating a substance at constant volume, the energy input only pays for the internal molecular kinetic energy.     * When heating at constant pressure, the substance expands. The energy input must pay for the internal temperature increase PLUS the "mechanical tax"—the work required to push the boundary against the surrounding environment.     * Helium Example (m=1kg,ΔT=1Cm = 1\,kg, \Delta T = 1\,^{\circ}C):         * Constant Volume heating: Requires 3.12kJ3.12\,kJ.         * Constant Pressure heating: Requires 5.19kJ5.19\,kJ.

  • Universality: The partial derivative equations for cpc_p and cvc_v are fundamental properties and are universally valid for any substance undergoing any process.

Ideal Gases: Properties and Relationships

  • Joule’s Observation: Joule demonstrated that when an ideal gas expands into a vacuum without heat transfer or work, its temperature does not change (T=CT = C).

  • Conclusions for Ideal Gases:     * Internal energy (uu), Enthalpy (hh), and specific heats (cv,cpc_v, c_p) are functions of temperature only (u=u(T)u = u(T), h=h(T)h = h(T), cv=cv(T)c_v = c_v(T), cp=cp(T)c_p = c_p(T)).     * They are independent of pressure (PP) and specific volume (vv).

  • Mathematical Relations:     * Building from the definition of enthalpy: h=u+Pvh = u + Pv.     * Substituting the Ideal Gas Law (Pv=RTPv = RT): h=u+RTh = u + RT.     * Differentiating the relationship: dh=du+RdTdh = du + R\,dT.     * Substituting dh=cpdTdh = c_p\,dT and du=cvdTdu = c_v\,dT yields the final relation: cp=cv+R(kJ/kgK)c_p = c_v + R\, (kJ/kg \cdot K).

  • Specific Heat Ratio (kk):     * Formula: k=cpcvk = \frac{c_p}{c_v}.     * For air at room temperature: k1.4k \approx 1.4.     * For monatomic gases: k1.667k \approx 1.667.

Frameworks for Calculating Internal Energy and Enthalpy Changes

There are three primary methods to determine Δu\Delta u and Δh\Delta h for ideal gases, ordered by accuracy and manual effort.

  • 1. Empirical Tables (The Gold Standard):     * Method: Uses pre-calculated, tabulated data based on a 0K0\,K reference state.     * Formula: Δu=u2u1\Delta u = u_2 - u_1.     * Pros: Highest accuracy; always preferred if available.

  • 2. Direct Integration:     * Method: Integrating the specific heat polynomial equations over the temperature range.     * Formula: Δu=T1T2cv(T)dT\Delta u = \int_{T_1}^{T_2} c_v(T)\,dT.     * Pros: Highly precise; ideal for computerized calculations.     * Cons: Mathematically tedious for manual work.

  • 3. Average Specific Heat (The Engineer’s Approximation):     * Method: Assuming a linear variation of specific heat over small temperature gaps and evaluating at the arithmetic mean temperature (TavgT_{avg}).     * Formula: Δucv,avgΔT\Delta u \cong c_{v,avg} \Delta T.     * Pros: Rapid approximation; fastest manual calculation.     * Cons: Less accurate, especially over large temperature intervals.

Practical Example: Air Heating at Constant Pressure

Scenario: Air at 300K300\,K and 200kPa200\,kPa is heated at constant pressure to 600K600\,K. Determine the change in internal energy per unit mass.

  • Assumptions: Air is considered an ideal gas (high temperature and low pressure relative to critical-point values).

  • Method (a): Air Table (Table A–17):     * u1=u@300K=214.07kJ/kgu_1 = u @ 300\,K = 214.07\,kJ/kg     * u2=u@600K=434.78kJ/kgu_2 = u @ 600\,K = 434.78\,kJ/kg     * Δu=u2u1=434.78214.07=220.71kJ/kg\Delta u = u_2 - u_1 = 434.78 - 214.07 = 220.71\,kJ/kg

  • Method (b): Functional Form / Integration (Table A–2c):     * Specific heat of air as a third-degree polynomial: cˉ<em>p(T)=a+bT+cT2+dT3\bar{c}<em>p(T) = a + bT + cT^2 + dT^3.     * Constants for air: a=28.11a = 28.11, b=0.1967×102b = 0.1967 \times 10^{-2}, c=0.4802×105c = 0.4802 \times 10^{-5}, d=1.966×109d = -1.966 \times 10^{-9}.     * Calculation: Δuˉ=</em>T1T2(cˉpRu)dT=6447kJ/kmol\Delta \bar{u} = \int</em>{T_1}^{T_2} (\bar{c}_p - R_u)\,dT = 6447\,kJ/kmol.     * On unit-mass basis: Δu=ΔuˉM=6447kJ/kmol28.97kg/kmol=222.5kJ/kg\Delta u = \frac{\Delta \bar{u}}{M} = \frac{6447\,kJ/kmol}{28.97\,kg/kmol} = 222.5\,kJ/kg.     * Error relative to table: 0.8%0.8\%.

  • Method (c): Average Specific Heat (Table A–2b):     * Average temperature (TavgT_{avg}) = 300+6002=450K\frac{300 + 600}{2} = 450\,K.     * cv,avgc_{v,avg} at 450K450\,K (from Table A-2b) = 0.733kJ/kgK0.733\,kJ/kg \cdot K.     * Δu=cv,avg(T2T1)=(0.733kJ/kgK)(600300)K=220kJ/kg\Delta u = c_{v,avg}(T_2 - T_1) = (0.733\,kJ/kg \cdot K)(600 - 300)K = 220\,kJ/kg.     * Error relative to table: 0.4%0.4\%.

  • Discussion: Accuracy decreases significantly if using cvc_v at the initial temperature (300K300\,K) instead of average temperature (215.4kJ/kg215.4\,kJ/kg, a 2%2\% error).

The Incompressible Model: Solids and Liquids

  • Definition: An incompressible substance is one whose specific volume (vv) or density remains constant during a process (v=constantv = constant).

  • Simplification: Because volume cannot change, the mechanical tax of expansion disappears. The distinction between heating at constant volume and constant pressure vanishes.     * cp=cv=cc_p = c_v = c     * Example: For Iron at 25C25\,^{\circ}C, c=0.45kJ/kgCc = 0.45\,kJ/kg \cdot ^{\circ}C.

  • Internal Energy Calculation:     * Primary equation: du=c(T)dTdu = c(T)\,dT.     * Integration: Δu=u2u1=T1T2c(T)dT\Delta u = u_2 - u_1 = \int_{T_1}^{T_2} c(T)\,dT.     * Standard approximation: ΔucavgΔT\Delta u \cong c_{avg} \Delta T.

  • Enthalpy Calculation:     * Start with definition: h=u+Pvh = u + Pv.     * Differentiate: dh=du+vdP+Pdvdh = du + v\,dP + P\,dv.     * Since volume is locked (dv=0dv = 0): dh=du+vdPdh = du + v\,dP, leading to Δh=Δu+vΔP\Delta h = \Delta u + v \Delta P.

  • Sub-Scenarios for Enthalpy:     * 1. Constant Pressure (e.g., Heaters): ΔP=0Δh=ΔucavgΔT\Delta P = 0 \rightarrow \Delta h = \Delta u \cong c_{avg} \Delta T. (Valid for all solids).     * 2. Constant Temperature (e.g., Pumps): ΔT=0Δu=0\Delta T = 0 \rightarrow \Delta u = 0. The equation collapses to Δh=vΔP\Delta h = v \Delta P. This is crucial for liquid fluid dynamics.

State Diagnosis and Master State Matrix

Correct thermodynamic analysis depends on diagnosing the state of the substance before selecting calculation tools.

Property

Ideal Gases (Compressible)

Solids & Liquids (Incompressible)

Volume Behavior

Variable (vv changes)

Constant (v=Cv = C)

Specific Heat Rule

cp=cv+Rc_p = c_v + R

cp=cv=cc_p = c_v = c

Internal Energy (Δu\Delta u)

u2u1u_2 - u_1 (Tables) or cv,avgΔTc_{v,avg} \Delta T

cavgΔTc_{avg} \Delta T

Enthalpy (Δh\Delta h)

h2h1h_2 - h_1 (Tables) or cp,avgΔTc_{p,avg} \Delta T

Δu+vΔP\Delta u + v \Delta P

Real Gas Behavior and Molecular Observations

  • Zero-Pressure Specific Heats: At low pressures, real gases approach ideal-gas behavior. Specific heats depend on temperature only, denoted as cp0c_{p0} and cv0c_{v0}.

  • Molecular Complexity: Specific heats of gases with complex molecules (two or more atoms) are higher and increase with temperature.

  • Linear Approximation: The variation of specific heats with temperature is smooth and can often be approximated as linear over intervals of a few hundred degrees or less.

  • Substitution: In many engineering applications, specific heat functions are replaced by constant average specific heat values for simplicity.