If two object with different temperatures are brought in to contact with one another thermal energy will flow from one to another until the temperatures are the same, and we then say that the objects are in thermal equilibrium.

The zeroth law of thermodynamics states that:

“If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.”

Most, but not all, materials expand when heated. The change in length of material due to linear thermal expansion is

$\Delta l=\alpha l_{0}\Delta T$

$\alpha$ is the coefficient of linear expansion of the material, measured in $\mathrm{(^{o}C)^{-1}}$

The length of the object after it's temperature has been changed by $\Delta T$ is

$l=l_{0}(1+\alpha\Delta T)$

A material expands in all directions, and if we are interested in the volume changes of a rectangular object, that is isotropic, meaning it expands in the same way in all directions, then

$\Delta V = \beta V_{0}\Delta T$

$V_{0}=l_{0}w_{0}h_{0}$ → $V=l_{0}(1+\alpha\Delta T)w_{0}(1+\alpha\Delta T)h_{0}(1+\alpha\Delta T)$

$\Delta V=V-V_{0}=V_{0}(1+\alpha\Delta T)^{3}-V_{0}=V_{0}[3(\alpha\Delta T)+3(\alpha\Delta T)^{2}+(\alpha\Delta T)^{3}]$

If $\alpha\Delta T << 1$ then $\beta \approx 3\alpha$

A mole of gas is a given number of molecules, Avagadro's number, $N_{A}=6.02\times 10^{23}$. If we have a certain mass $m$ of a gas which has a certain molecular mass (measured in atomic mass units, $\mathrm{u}$, which are also the number of grams per mole.), the the number of moles $n$ is given by

$n=\frac{m[\mathrm{g}]}{\textrm{molecular mass}[\mathrm{g/mol}]}$

and

$PV=nRT$ where $R=8.314\mathrm{J/(mol.K)}$

The ideal gas law can also be written in terms of the number of molecules

$PV=nRT=\frac{N}{N_{A}}RT=NkT$

where $k=\frac{R}{N_{A}}=\frac{8.314\mathrm{J/(mol.K)}}{6.02\times 10^{23}}=1.38\times 10^{-23}\mathrm{J/K}$ is the Boltzmann Constant.

The average kinetic energy of molecules in an monatomic ideal gas is

$\bar{K}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$

$f(v)=4\pi (\frac{m}{2\pi k T})^{\frac{3}{2}}v^{2}e^{-\frac{1}{2}\frac{mv^{2}}{kT}}$

Conduction-Primary mechanism for solids in thermal contact with each other. The heat flow $\Delta Q$ during a time interval $\Delta t$ in a conductor of length $l$ and area $A$ which connects two object's which have temperature $T_{1}$ and $T_{2}$ is

$\frac{\Delta Q}{\Delta t}=kA\frac{T_{1}-T_{2}}{l}$

which in differential form is

$\frac{dQ}{dt}=-kA\frac{dT}{dx}$

Convection-Movements of molecules in a gas or liquid.

Radiation-Electromagnetic transmission of heat, does not require a medium.

$\frac{\Delta W}{\Delta t}=\epsilon\sigma A T^{4}$

$\sigma=5.67\times10^{-8}W/m^{2}K^4$ and $\epsilon$ is the emissivity of the surface, a perfect surface for emission or asborption (a black surface) has an emissivity of 1, whereas a shiny surface that neither absorbs or transmits would have an emissivity of zero. Most materials are somewhere in between these two limits.

A quantity of heat, $Q$, flowing into an object leads to a change in the temperature of the object, $\Delta T$, which is proportional to it's mass $m$ and a characteristic quantity of the material, it's specific heat, $c$

$Q=mc\Delta T$

We can see that heat flowing in to an object is positive $\Delta T>0$ and heat flowing out is negative $\Delta T < 0$

The specific heat is the heat capacity per a unit of mass, in SI the units of specific heat are $\mathrm{\frac{J}{kg.K}}$.

The assumption of an isolated system is very useful in problem solving as it says that the sum of the heat transfers in the system must be zero.

$\Sigma Q = 0$

In a system where the different objects start at different temperatures, but eventually come to an equilibrium temperature $T$

$\Sigma Q = m_{1}c_{1}(T-T_{i1})+m_{2}c_{2}(T-T_{i2})+..$

Phase changes from a low temperature phases to a high temperature phase require a certain amount of heat, called the latent heat.

The latent heat of of fusion, $L_{f}$, refers to a change from solid to liquid and the latent heat of vaporization, $L_{v}$, refers to a change from liquid to gas. The heat required to change a mass $m$ of a substance from one phase to another is

$Q=mL$

During a change from one phase to another the temperature of the system remains constant.

The first law of thermodynamics, dictates how internal energy, heat and work are related to each other. For a closed system the first law states that the change in the internal energy of a system, $\Delta E_{int}$, is the sum of the heat added **to** the system $Q$ and the net work done **by** the system $W$.

$\Delta E_{int}=Q-W$

The table shows some of the results that apply to a particular kind of thermal process

Process | Constant | ΔE_{int} | Q | W |
---|---|---|---|---|

Isothermal | T | 0 | Q=W | W=Q |

Isobaric | P | Q-PΔV | ΔE_{int}+PΔV | PΔV |

Isovolumetric | V | Q | ΔE_{int} | 0 |

Adiabatic | -W | 0 | -ΔE_{int} |

For an ideal gas $P=\frac{nRT}{V}$, so for an isothermal process

$W=\int_{V_{A}}^{V_{B}}P\,dV=nRT\int_{V_{A}}^{V_{B}}\frac{dV}{V}=nRT\ln\frac{V_{B}}{V_{A}}$

In an isobaric process the pressure is constant so the work is

$W=\int_{V_{A}}^{V_{B}}P\,dV=P(V_{B}-V_{A})=P\Delta V$

and if the system is an ideal gas

$W=P(V_{B}-V_{A})=nRT_{B}(1-\frac{V_{A}}{V_{B}})=nRT_{A}(\frac{V_{B}}{V_{A}}-1)$

In an isovolumetric process the work done is zero

$W=0$

For an ideal gas $c_{P,m}-c_{V,m}=R$

The equipartition theorem can be used to explain the higher heat capacity of more complicated gases. The equipartition theorem states that energy is equally shared between the different degrees of freedom the molecules in the gas have.

Each degree of translational or rotational freedom contributes $\frac{1}{2}R$ to the molar specific heat at constant volume $c_{V,m}$

$PV^{\gamma}=\mathrm{constant}$

**Heat cannot spontaneously flow from a cold object to a hot one, whereas the reverse, a spontaneous flow of heat from a hot object to a cold one, is possible.**

or

**No device is possible whose sole effect is to transform a given amount of heat directly in to work.**

$Q_{H}=W+Q_{L}$

$e=\frac{W}{Q_{H}}$

$e=\frac{W}{Q_{H}}=\frac{Q_{H}-Q_{L}}{Q_{H}}=1-\frac{Q_{L}}{Q_{H}}$

We will define the change in entropy in a reversible process at constant temperature as

$\Delta S =\frac{Q}{T}$

If we want to treat non-constant temperature cases we can express the change of entropy in differential form

$dS=\frac{dQ}{T}$

and then the change in entropy in going from state $a$ to $b$ will be

$\Delta S =S_{b}-S_{a}=\int_{a}^{b}\,dS=\int_{a}^{b}\frac{dQ}{T}$