19.1 Heat as Energy Transfer #
Units #
 Heat unit is calorie (cal)
 The amount of heat needed to raise the temperature of 1 gram of water by 1 celsius
 $$4.186 \text{ J} = 1 \text{ cal}$$
 Kilocalorie (kcal, Calorie) is more common
 Amount of heat needed to raise 1 kg of water by 1 celsius
 $$4.186 \text{ kJ} = 1 \text{ kcal}$$
 British system of units has British thermal units (Btu)
 One Btu is the heat needed to raise the temperature of 1 lb of water by 1 Fahrenheit
 $$1 \text{ Btu} = 0.252 \text{ kcal} = 1056 \text{ J}$$
 Gas companies use the unit therm: .$10^5 \text{ Btu}$
Heat #
 Heat is energy transferred from one object to another because of a difference in temperature.
 Energy transfers from hot to cold object until equilibrium
 The SI units for heat is the joule: this is because heat is a form of energy!
19.2 Internal Energy #
 Internal Energy: The sum of all the energy of all the molecules in an object
 Sometimes called thermal energy
Difference between Temp, Heat, and Internal Energy #
 Temperature is the average kinetic energy of all of the molecules
 Internal energy is the sum of the energy of all of the molecules
 E.x. Two equalmass iron ingots could the same temperature as a single ingot, but the two would have double the internal energy
 Heat refers to the transfer of energy from one object to another due to a difference in temperatures
 Direction of transfer depends on temperature, not internal energy
 E.x. .$50\text{ g}$ of .$30^\circ\text{ C}$ water mixed with .$200 \text{ g}$ of .$25^\circ \text{ C}$ water results with heat transferring from the smaller sample with less internal energy to the larger sample with more internal energy.
Calculating Internal Energy #
 Internal energy is the sum of all the translational kinetic energy of the molecules in a monatomic gas
 Monatomic: Gas with one atom per molecule
 We can rewrite this as the average KE per molecule times the total number of molecules, .$N$ $$E_{\text{int}} = N \bigg(\frac{1}{2}m\bar{v}^2\bigg) = \frac{3}{2}Nk_B T = \frac{3}{2}nRT$$
 We can see that internal energy for a monatomic gas depends only on the temperature and number of moles
 If a gas isn’t monatomic, then we need to consider the rotational and vibrational energy of the molecules
 Nonmonatomic gasses result in a internal energy at a given temperature compared to a monatomic
 The internal energy of real gases depends mainly on temperature
 There are some exceptions of gases depending on pressure and volume as well
 Internal energy of liquids and solids is more complex
 It includes electric potential energy of the chemical bonds
19.3 Specific Heat #
 Amount of heat required to change the temperature of a material is found with the following: $$\Delta Q = mc \Delta T$$
 .$c$ specific heat capacity that depends on the material .$[\text{J}/(\text{C}^\circ\text{ kg})]$
 For water at .$15 ^\circ \text{ C}$ and constant pressure .$1 \text{atm}$, .$c = 4168 \text{ J}/(\text{C}^\circ\text{ kg}) = 1.00 \text{ kcal}/(\text{C}^\circ \text{ kg})$
 .$c$ does vary to some extent with temperature (and slightly pressure), but for small .$\Delta T$ we can say .$c$ is a constant
 Relative to other materials/substances, water has a high specific heat capacity
19.4 Calorimetry #
Types of Systems #
 System: Any (set of) object(s) we choose to consider
 Closed System: Mass is constant, but energy may be exchanged within environment
 Isolated: If no energy in any form passes across its boundaries
 We idealize systems to be closed systems, which is rare in the real world
 Heat will flow from hot to cold region of system until equilibrium
 We can assume that no energy is lost; heat lost in one part = heat gained in another part or .$\Sigma Q = 0$
 Open System: Mass and energy may enter/leave
Calorimeter #
 Calorimetry: Quantitative measure of heat exchange
 Calorimeter tend to have insulation so that no heat is exchanged with the surrounding air
 Often use thermometer to measure change the temperature
 E.x. a substance sample will be heated up, measured, then quickly placed inside cool water of calorimeter
 The heat lost from the sample will be gained by the water and the calorimeter cup
 Measuring final temperature of the mixture lets us calculate the specific heat
 Assume that small masses like the thermometer/stirrer are negligible
19.5 Latent Heat #
 Change of Phase: When a material changes from solid to liquid or liquid to gas.
 A certain energy is required for a phase change
 During phase changes, temperature stops increasing and all energy goes into the phase change
 Latent heat is lost during phase change (often in the form of heat)
 Heat of fusion
 .$L_F$: Heat required to change .$1.0 \text{ kg}$ of a substance from solid to liquid state
 Heat fusion of water is .$79.7 \text{kcal/kg} = 333 \text{kJ/kg}$
 Heat of Vaporization
 .$L_V$: Heat required to change a substance from liquid to vapor phase
 Heat vaporization is .$539 \text{kcal/kg} = 2260 \text{kJ/kg}$
 Heat involved in the phase change depends on the mass and latent heat: $$\Delta Q = mL$$
 Therefore, when considering the change in a system involving heating a substance to a phase change (e.g. boiling at temperature .$T$), we can write: $$\Delta Q_{\text{total}} = mc \Delta T + mL$$
Evaporation #
 Heat of Vaporization of water increases slightly with a decrease in Temperature
 At .$20^\circ \text{ C}$, it’s .$585 \text{ \text{kcal/kg}}$
 When liquid evaporates, the remaining liquid cools because the heat/energy comes from the water itself
 Therefore, internal energy decreases with evaporation
Kinetic Theory of Latent Heats #
 At melting point, the latent heat of fusion doesn’t increase the average KE / temperature
 Rather, the energy goes into overcoming the PE associated with the forces between the molecules
 Once the molecules in a solid are broken from there lattice formation, they can freely roll over one another as a liquid
 More energy is required for liquid to gas phase because the average distance between the molecules is greatly increased
 The larger the distance that the molecules have to be separated, the more work has to be done to pull them apart
19.6 First Law of Thermo #
 Heat and work are different
 Heat is the transfer of energy due to a difference in temperature – hot/cold bath around gas chamber
 Work is the transfer of energy not due to a temperature difference – piston applying force to a gas
 Internal energy and temperature are both proportional to heat and work though with the following equation: $$\Delta E_{\text{int}} = Q  W = E_{\text{int, 2}}  E_{\text{int, 1}} \ \ \ \text{[First Law of Thermo.]}$$
 .$Q$ is net heat added to the system
 Heat added is .$\texttt{+}$, heat lost is .$\texttt{}$,
 E.x. gas chamber heated .$\Longrightarrow$ system gains energy
 .$W$ is net work done by the system
 Work done by system is .$\texttt{+}$, work done on the system is .$\texttt{}$,
 E.x. gas expands .$\Longrightarrow$ system looses energy
 .$Q$ and .$W$ are not state variables in that a static state doesn’t have “heat” or “work” – only when the system changes through thermodynamic process can we measure heat/work.
 This is unlike .$P, V, T$ and .$E_{\text{int}}$ which are **state variables** (can be measured at all states)
 We can also extend the first law to include systems that have KE and PE: $$\Delta K + \Delta U + \Delta E_{\text{int}} = Q  W = E_{\text{int, 2}}  E_{\text{int, 1}}$$
19.7 Thermodynamic Process and the 1st Law #
Isothermal Process (.$\Delta T = 0$) #
 When temperature is constant, .$PV = \text{constant}$
 Each label of points in the graph above represent the systems states (it’s pressure and temperatures)
 Isotherms: curves in PV diagram
 At a lower temperature, an isothermal process would be represented by the isotherm .$A’B'$
 We also assume that the container is a heat reservoir: a body whose mass is so big that the temperature doesn’t change when heat is exchanged
 We increase internal energy by doing work, such as by decreasing the volume of the container with by applying a force to a piston over some distance
 We assume that expansion/compression is quasistatic: we decrease the volume slow enough that we can consider it a series of equilibrium states all at the same temperature
 E.x. if we started with state .$A$ and added heat .$Q$ to the system, the system would reach point .$B$
 If .$T$ remain constant, the volume will expand, both doing work .$W$ on the environment and decreasing the .$P$
 We know .$E_\text{int} = \frac{3}{2}nR\Delta T$, and since .$\Delta T = 0 \Longrightarrow E_\text{int} = 0$
 Thus, .$E_\text{int} = Q  W \Longrightarrow W = Q$
Adiabatic Process (.$Q = 0$) #
 No heat allowed to flow in our out of system. This can happen if…
 Process happens so quickly that heat, a slow process, has no time to flow in/out
 E.x. a combustion engine happens quickly it’s nearly adiabatic
 System is well insulated
 Process happens so quickly that heat, a slow process, has no time to flow in/out
 If a system experiences an adiabatic process slowly, it will look similar to curve .$AC$
 Since .$Q = 0 \Longrightarrow \Delta E_\text{int} = W$
 In a reverse processes represented by .$CA$ (adiabatic compression), work is done on the gas so .$E_\text{int}$ and .$T$ rise
Isobaric and Isovolumetric #
 Isobaric: .$\Delta P = 0 \Longrightarrow Q = \Delta E_\text{int} + W = \Delta E_\text{int} + P\Delta V$. The heat transferred to the system does work, but also changes the internal energy of the system
 Isovolumetric: .$\Delta V = 0 \Longrightarrow W = 0 \Longrightarrow Q = \Delta E_\text{int}$. The thermodynamic process is the addition or removal of heat.
 First law of thermo holds for both of these processes
Work done in volume changes #
 For quasistatic processes: $$dW = \vec{F} \cdot d\vec{l} = PA d\vec{l} = P\ dV \ \ \ \text{(1)}$$ $$W = \int dW = \int_{V_A}^{V_B} P\ dV \ \ \ \text{(2)}$$
 .$\text{(1)}$ Where .$F = PA$ is the force the gas exerts on the piston and .$d\vec{l}$ is the (small) distance the piston moves
 .$\text{(2)}$ This shows that the work done is the area under the .$PV$ curve
 This equations are valid for work done in any volume change (solid, liquids, gas)
 .$W$ (and even .$Q$) depends on the initial and final states and also on the process (or path)
Isothermal #
Isovol. and Isobaric #
$$W_{\text{Isovol.}} = 0$$ $$W_{\text{Isobaric}} = \int_{V_A}^{V_B} P \ dV$$ $$… = P \Delta V$$ $$… = P_B(V_B  V_A)$$ $$… = = nRT_B(1  \frac{V_A}{V_B})$$
Free Expansion #
 A type of adiabatic process where gas is allowed to expand in a volume without doing any work
 Must be done with insulated containers so that no heat is able to flow in/out; .$Q = 0$
 No work is done either because no object is moved; .$W = 0$
 Thus, .$\Delta E_\text{int} = 0$
 In reality, we see temperature slightly drops meaning internal energy does depend on pressure or volume as well as temperature.
19.8 Molar Specific Heats for Gases and Equipartition of Energy #
Molar Specific Heat #
 Specific heat for gases depends heavily on the process and how it’s carried out
 Specific heat for constant pressure and constant volume vary
 We use molar specific heat for gases: .$C_V$ and .$C_P$ which are defined as the heat required to raise .$1 \text{ mol}$ of gas by .$1^\circ \text{ C}$ at a constant volume or pressure respectively.
 We then use .$n$ instead of .$m$ in our heat equations: $$\Delta Q = nC_V \Delta T\ \ \ \text{[Constant Volume]}$$ $$\Delta Q = nC_P \Delta T\ \ \ \text{[Constant Pressure]}$$ which we can then relate to the specific heat with .$M$ as the molecular mass of the gas, .$m/n$ in grams/mol: $$C_V = Mc_V$$ $$C_p = Mc_p$$
 In a heating process, when .$\Delta V = 0$ then the heat added, .$Q_V$ goes entirely into internal energy: .$Q_V = \Delta E_\text{int}$
 However, when pressure is constant work is done. Thus, heat added, .$Q_P$, goes towards increasing internal energy and work: .$W=P\Delta V$
 Therefore, more heat is needed for a constant pressure system: .$Q_P = \Delta E_\text{int} + P\Delta V$
 Since .$\Delta E_\text{int}$ is the same for both processes, we can write .$Q_P  Q_V = P \Delta V$
 With an ideal gas, we know .$V = nRT/P$ so .$\Delta V = nR\Delta T/P$ which we can combine with the prior equations to get: $$nC_P\Delta T  nC_V \Delta T = P\bigg(\frac{nR\Delta T}{P}\bigg) \Longrightarrow C_P  C_V = R$$
 We can also relate internal energy to molar specific heat for monatomic gases at constant volumes: $$\Delta E_\text{int} = Q_V \Longrightarrow \frac{3}{2}nRT = nC_V\Delta T \Longrightarrow C_V = \frac{3}{2}R$$
 We can then plug in our new value for .$C_V$ into the second to last equation .$C_P  C_V = R$ to get .$C_P = \frac{5}{2}R$ for a monatomic gas.
Equipartition of Energy #
 Degrees of Freedom: The number of independent ways a molecule can posses energy
 Degrees of freedom depend on the temperature
 At low temperatures, the only degree of freedom is from translational .$KE$
 Starting after .$0K$
 Diatomic gas: .$C_V = \frac{3}{2}R$ (3 for each axis)
 Sum of .$\frac{1}{2}m \langle v_x, v_y, v_z \rangle$
 At “regular” temperatures, the molecules posses rotation energy
 Around .$50K$
 Diatomic gas: .$C_V = \frac{5}{2}R$
 Sum of .$\frac{1}{2}I \langle 0, \omega_y, \omega_z \rangle$ (since it’s rotating about .$\hat x$ meaning .$E_{\text{rotational, }x}) = 0$
 At higher temperatures, the molecules gain energy associated with their vibrations:
 Around .$1000K$
 One from KE of the molecules vibrating back and forth: .$\frac{1}{2}mv_{\text{COM}}^2$
 The second from PE of the vibrational motion (think of this as a spring’s PE): .$\frac{1}{2}kx^2$
 Solids:
 The molar temperature of solids at high temperatures is close to .$3R$.
 At high temperatures, there are six degrees of freedom: three from vibrational KE in the .$x, y,$ and .$z$ axis and three more from spring PE in the same axis
 Some of these degrees of freedom aren’t active at lower temperatures
 Principle of Equipartition of Energy: Energy is shared equally among degrees of freedom and each degrees has energy .$\frac{1}{2}k_B T$
 Thus, for a particle with three degrees of freedom (such as a monatomic gas) .$C_V = \frac{3}{2}R$
 Diatomic gases have five degrees so they have .$C_V = \frac{5}{2}R = 4.97 \text{ cal/(mol K)}$ and have .$E_\text{int} = N(\frac{5}{2}k_B T) = n C_V \Delta T = \frac{5}{2}nRT$ where .$n$ is the number of moles and .$N$ is the number of molecules
19.9 Adiabatic Expansion of a Gas #
 The .$PV$ curve for adiabatic expansion (.$Q=0$) is slightly less steep than isothermal processes (.$\Delta T = 0$)
 This means that for the same change in volume, the pressure will be greater in adiabatic processes
 Therefore, the temperature of a gas must drop in adiabatic expansion and rise in adiabatic compression
 We can relate .$P$ and .$V$ for a quasistatic expansion / compression with $$PV^\gamma = \text{[constant] for } \gamma = \frac{C_P}{C_V}$$
19.10 Heat Transfer #
Conduction #
 Conduction can be visualized thinking of molecular collisions
 The hot end of an object has fast moving molecules
 These molecules bump into other molecules, transferring them some of their own KE
 This keeps repeating down the object
 Free electrons are the primary source of these collisions
 Heat conduction only occurs when there is a difference in temperatures
 Heat conduction rate is proportional to the difference in temperatures: $$\frac{Q}{t} = kA\frac{T_1  T_2}{l}$$
 Where .$A$ is the cross section area, .$l$ is the distance between the two ends, and .$k$ is a constant called thermal conductivity that depends on the material
 Good insulator have a low .$k$
 Building materials sometimes list the thermal resistance, .$R$, which is equal to .$R = \frac{l}{k}$ where .$l$ is the material’s thickness
 Larger .$R$ means better insulation
 If .$k$ or .$A$ isn’t constant, we consider a small thickness: $$ \frac{dQ}{dt} = kA \frac{dT}{dx}$$
 .$\frac{T_1  T_2}{l} \text{and} \frac{dT}{dx}$ is called the temperature gradient
 We have a negative sign in the equation above because the direction of heat flow is opposite to the temperature gradient
Convection #
 Convection involves heat flowing by the bulk movement of molecules from one place to another
 Whereas conduction involved molecules/electrons moving over small distances, convection involves the movement of a large number of molecules over a long distance
 Natural Convection occurs in systems where a cold substance (air, water) is warmed and subsequently expands, decreasing density and thus rising
Radiation #

Whereas conduction and convection require a medium, radiation doesn’t
 The sun’s rays are a form of heat and travel through (nearly empty) space
 Radiation of the sun’s rays arrive on a clear day at a rate around .$1000 \text{W/m}^2$
 Most of the time radiation consists of electromagnetic waves, but infrared (IR) wavelengths are responsible for heating Earth
 The sun’s rays are a form of heat and travel through (nearly empty) space

The rate at which energy leaves a radiation object, .$Q/t$, is $$ \frac{Q}{t} \epsilon \sigma A T^4$$

.$\epsilon$ is called emissivity.
 Between 0 and 1
 Characteristic of the surface of the radiating material
 Black surfaces close to one, shiny metal surfaces close to zero
 Depends slightly on the temperature of the material
 A good absorber is also a good emitter
 A black tee shirt gets very hot because it absorbs nearly all the radiation that hits it

.$\sigma$ is the StefanBoltzmann constant: .$\sigma = 5.67 \cdot 10^{8} \text{ W/(m}^2 \text{K}^4\text{)}$

Objects also absorb heat of surrounding objects. This net heat flow can be found by $$ \frac{Q}{t} \epsilon \sigma A (T_1^4  T_2^4)$$

Where .$T_1$ is the object’s temperature and .$T_2$ is the surrounding environment’s temperature

.$T_1 > T_2$: net flow of heat is from object to the surroundings

.$T_1 < T_2$: net flow of heat is from surroundings into object, raising the object temperature