Specific Heat Capacity

Specific Heat Capacity

• Different substances need varying amounts of energy to raise the temperature of identical masses by one degree
• For example, raising the temperature of 1 kg of iron from 20 to 30 degrees Celsius requires 4.5 kJ of energy
• Raising the temperature of 1 kg of liquid water by the same amount requires about 9 times more energy, specifically 41.8 kJ
• Specific heat is the energy to raise the temperature of a unit mass of a subject by one degree in a specific way
• Thermodynamics focuses on specific heat at constant volume (Cv) and specific heat at constant pressure (Cp)
• Cv refers to the energy needed to raise the temperature of a unit mass of a substance by one degree while keeping the volume constant
• Cp refers to the energy needed to achieve the same temperature increase while keeping the pressure constant

Specific Heats in Terms of Thermodynamic Properties

• For a fixed mass in a stationary closed system undergoing a constant-volume process, the conservation of energy principle is: 𝛿𝑒𝑖𝑛 − 𝛿𝑒𝑜𝑢𝑡 = 𝑑𝑢
• 𝛿𝑒𝑖𝑛 − 𝛿𝑒𝑜𝑢𝑡 represents the net amount of energy transferred to the system
• Based on the definition of Cv, the energy being transferred must be equal to Cv dT, with dT representing the differential change in temperature
• Therefore Cv = (𝛿𝑒𝑖𝑛 − 𝛿𝑒𝑜𝑢𝑡)/dT, implying 𝛿𝑒𝑖𝑛 − 𝛿𝑒𝑜𝑢𝑡 = 𝐶𝑣 𝑑𝑇, at constant volume also expressed as 𝑪𝒗 𝒅𝑻 = 𝒅𝒖
• Cv represents the change in internal energy with temperature at constant volume
• An expression for Cp can be derived by considering a constant-pressure expansion or compression process: 𝐶𝑝 = (𝜕ℎ/𝜕𝑇)p
• Cp represents the change in enthalpy with temperature at constant pressure
• Cv is defined as the change in internal energy of a substance per unit change in temperature at constant volume
• Cp is defined as the change in the enthalpy of a substance per unit change in temperature at constant pressure
• These equations are property relations and apply to any process and are valid for any substance
• Cv relates to changes in internal energy, while Cp relates to changes in enthalpy

Enthalpy

• Enthalpy refers to the total heat content in a system
• Under constant pressure conditions, with volume (V) and temperature (T) changing, there is a change in internal energy (ΔU)
• δQ = dU + δW, where δQ is heat transfer, dU is change in internal energy, and δW is work done can be expressed as 𝑄1−2 = 𝑈2 − 𝑈1 + 𝑃 𝑉2 − 𝑉1
• Enthalpy 𝑄1−2 = 𝐻2 − 𝐻1, or 𝑸1−2 = ∆𝐇 can be calculated as H= U + PV

Temperature Dependency

• In 1843 Joule submerged two tanks connected with a pipe and valve in a water bath
• Initially, one tank had high pressure air and the other was evacuated
• Following thermal equilibrium, he opened them to allow air between them
• Joule noticed no temperature changes in the water bath, assuming no heat transfer to or from the air
• Since there was no work done, joule deduced no changes in internal air energy happened although volume and pressure changed
• It was reasoned that internal energy relies on temperature and is not based on pressure or volume
• Internal energy (u) is a function of temperature (T): u = u(T)
• Given h = u + PV and PV = RT, it follows that h = u + RT and h = h(T)
• Changes to internal energy and enthalpy are calculated using du = Cv(T)dT and dh = Cp(T)dT
• Where ∆u = ∫ 𝐶𝑣𝑇𝑑𝑇 and ∆h = ∫ 𝐶𝑝𝑇𝑑𝑇
• At low pressures, real gases behave like ideal gases, so specific heats depend only on temperature
• Specific heats of real gases at low pressures are called ideal-gas specific heats, denoted as Cpo and Cvo

Ideal Gas Considerations

• These equations are valid for any ideal gas undergoing any process: u2-u1= Cv,avg(T2-T1) and Δu =Cv ΔT with h2-h1= Cp,avg(T2-T1) and Δh =Cp ΔT

Energy Balance for closed system

• Closed system refers to the control mass where only energy can transfer in and out of the system
• In a closed system, increase in the energy of a potato in a oven = the amount of heat transferred to it (if work is zero)
• In a closed system, in the absence of work interactions, the energy change is equal to the net heat transfer (meaning work is zero)

Work in an Adiabatic System

• In a closed system, the electrical work in an adiabatic system is equal to the increase in the system's energy
• In a closed system, shaft work done in an adiabatic system is equal to the increase in the system's energy

Energy Change in a Closed System

• Energy change is the energy at the final state subtracted from the energy at the initial state
• For a system, ΔEsystem = Efinal - Einitial = E2 – E1 refers to total net work and heat transfer between the system and surroundings
• Energy can be internal (sensible, latent, chemical, nuclear), kinetic, potential, electric, or magnetic
• The sum of energies constitutes the systems total energy E with, ΔE = ΔU + ΔKE + ΔPE
• Where changes to internal energy, kinetic energy and potential energy are calculated as follows: ΔU = m(u2 – u1), ΔKE = 1/2 mV22 – V12), ΔPE = mg(z1 – z2)
• For stationary systems, ΔKE = Δ PE = 0 so ΔE = ΔU

Energy Transfer

• Energy transfer mechanisms include heat transfer, work transfer, and mass flow
• Where Ein – Eout = (Qin – Qout) + (Win – Wout) + (Emass, in – Emass, out) = ΔEsystem, E refers to energy
• For a closed mass system, the mass is constant and Ein – Eout = (Qin – Qout) + (Win – Wout) = ΔE
• In the case of Qin>Qout and Wout> Win then (Qnet, in Wnet, out) = ΔE general form of energy in KJ
• ΔE is equal to Q− W in units of Kj per sec (watt)
• q - w = Δ e in terms of Kj per Kg
• δq - δw = d is expressed in differential form

Cyclic Process

• A process is called a cycle if the initial and final states are the same, this process can not be isolated
• A cyclic process appears as a closed curve is a PV diagram
• The internal energy is zero since the state variable is zero, for a cycle, ΔE = 0, this means Q = W

Examples

• In a cooling of a hot fluid in a tank example from Yunus A. Cengel 8th edition, a rigid tank contains fluid cooled whilst stirred by a paddle wheel
• Initially the internal fluid energy = 800KJ, during cooling the fluid looses 500KJ of heat whilst the paddle wheel does 100KJ of work on the fluid
• The final energy of the fluid is found by neglecting paddle wheel energy
• The Boundary Work for a Constant-Pressure Process from Yunus A.Cengel 8th Edition shows a frictionless piston cylinder with 10lbm of steam at 60psia at 320F
• When heat is transferred to the steam to reach 400F degrees with a constant piston mass, the work is done by the steam during the process where P0 = 60 psia
• The Isothermal Compression of an Ideal Gas example from Yunus A.Cengel 8th Edition, demonstrates a piston initially containing 0.4m3 of air at 100kpa and 80C
• The air is now compressed to 0.1m3 whilst the cylinder stays constant, the work done can be determined during the process where T0 = 80 degrees Celsius
• For example 2.6 isothermal work from J.M. Smith is where horizontal piston cylinder arrangement is placed in a constant temperature bath with low piston frictions
• When an external force hold against an initial pressure of 14 bar with an initial volume of 0.03m3 while the product PV is constant
• The question asks what work is done if the external forces were used suddenly instead of gradually being reduced?

Homework

• Example 2-1 JM Smith 7th edition asks what is the energy of a waterfall when water flows (1KG) over a 100m waterfall without and energy change in the surrounding
• It asks what are the potential and kinetic energies related to base and bottom falls and any state changes
• Example 2-11 acceleration air by fan where fan uses electricity (20W) and transfers ventilation in room (1.0 kg/s at 8m/s), is this reasonable?