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Chapter 21 - Heat and the First Law of Thermodynamics

What is heat?

Heat should not be confused with temperature.

Heat is the energy transferred from one body to another due to thermal contact when the bodies are at different temperatures.

An early theory of heat was the caloric theory, which suggested that there was a fluid called caloric that was transferred from a body with high or low temperature. The idea that there was a particular substance associated with heat would imply conservation of heat.

A key player in overturning the caloric theory was James Prescott Joule. Through a number of experiments, initially motivated by his interest in using electric motors to power his brewery he was able to the equivalence of energy and heat.

Units of heat

As heat is a form of energy it can be measured in Joules (which is the SI unit for heat).

However its is very common to measure it in calories

$1 \mathrm{cal}=4.186 \mathrm{J}$

A calorie $(\mathrm{cal})$ is the amount of heat necessary to raise the temperature of 1 gram of water by 1 $\mathrm{^{o}C}$.

Energy in food is often measured in kilocalories $(\mathrm{kcal})$, which, somewhat confusingly, are refered to as Calories.

Another unit of heat, ironically most commonly used in the United States, is the British thermal units or BTU. The exact value of a BTU depends on where you are and how warm it is…but we can take it to be 1054.8 J.

Heat transfer

Heat transfer occurs by several different mechanisms.

  • Conduction-Primary mechanism for solids in thermal contact with each other.
  • Convection-Movements of molecules in a gas or liquid.
  • Radiation-Electromagnetic transmission of heat, does not require a medium.

Conduction

Conduction occurs by neighboring particles transferring energy from one to another. The ability of a material to conduct heat is measured by a parameter called the thermal conductivity, $k$. In the case of metal much of the conduction occurs through the free electrons which are also responsible for the electronic conduction. Non-metallic solids can also be good conductors of heat, lattice vibrations can also be an extremely effective way of transferring energy from one part of a material to another.

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

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

When dealing with practical situations the thermal conduction properties of a specific piece of building material will often be given as a thermal resistance $R$ where

$R=\frac{l}{k}$

and $l$ is the thickness of the piece of the material.

Convection

Convective heat transfer is heat transfer by the bulk motion of fluid.

Natural convection currents can occur due to changes in density as the temperature of a fluid changes.

Convection can also be forced, by the use of fans, stirrers or pumps.

Radiation

Thermal energy within a material is converted in to electromagnetic radiation.

The rate of radiation leaving the surface of a material with area A is given by the Stefan-Boltzmann equation.

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

A body also absorbs radiation from it's surroundings with temperature $T_{s}$ according to

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

$\sigma$ is the Stefan-Boltzmann constant

$\sigma=5.67\times10^{-8}W/m^{2}K^4$ 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.

If the body is in thermal equilibrium with it's surroundings, then $T=T_{s}$, the rate of emission equals the rate of absorption and the rate of heat flow is zero.

Specific Heat Capacity

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 specific heat of a material can depend on the conditions, for example the table in your textbook gives specific heats for materials at a fixed pressure per unit mass, the table in wikipedia gives these and also molar specific heats at constant pressure/constant volume. We will look at the effect of changes in volume on heat capacity by using the first law of thermodynamics in our next lecture.

Systems

When considering a set objects in a calorimetry problem we need to consider the boundary conditions on the system.

If a system has constant mass, with none either being lost or added, then we can say it is a closed system. This does not necessarily mean that energy cannot enter of leave. A system in which the total amount of energy is conserved is called an isolated system. A system in which both mass and energy can enter or leave is called an open system.

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 $T$

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

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Latent Heat

To this point we have considered systems where the constituents all stay in the same phase. 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. A good example we know from our everyday experience involves heating water from ice. By adding ice to water we quickly have the temperature of the solution equilibrate to $0\mathrm{^{o}C}$ and stay there as long as some ice remains.

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First Law of Thermodynamics

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 first law is a powerful statement of the conservation of energy, indeed conservation of energy can only be understood once we understand that heat is a transfer of energy.

Isothermal Processes

An isothermal process is one which occurs at a constant temperature. In order for such a process to occur there should be a heat reservoir, a large body in thermal contact with the system which has a temperature that will not significantly change as heat is exchanged with it. We will consider a quasistatic process, which means it occurs slowly enough that the system can be considered to be in equilibrium at any given time.

From the viewpoint of the first law

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

$\Delta E_{int}=\frac{3}{2}nR\Delta T = 0$

Which implies that the work done is equal to the heat added

$W=Q$

Adiabatic Process

In an adiabatic process no heat is allowed to flow in to or out of a system so $Q=0$. This can either be achieved by very good thermal insulation or in a process which occurs very quickly, so that there is no time for heat flow to occur.

In this case

$\Delta E_{int}=-W$

Remember that $W$ is the work done by the system, so when a gas expands it does work and the internal energy, and hence the temperature of the gas decreases. When a gas is compressed, work is done on it so it's temperature increases.

Isobaric and Isovolumetric processes

We can also consider processes that occur at constant pressure, referred to as isobaric processes, and those which occur at constant volume, referred to as isovolumetric or isochoric.

Work done in changing the volume of a gas

In order to apply the First Law to gases we should consider the amount of work done when a gas changes it's volume.

If we consider a piston full of gas at pressure $P$ the force on the piston head of area $A$ is $F=PA$. The work done in moving the head through an infinitesimal displacement $dl$ is

$dW=F\,dl=PA\,dl=P\,dV$

Where $dV$ is the change in volume

So the work done in going from one volume to another is

$W=\int dW=\int_{V_{A}}^{V_{B}}P\,dV$

The work done therefore depends on how the pressure varies during the change in the volume of the gas.

Path dependence of the work

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

$W=0$

Graphically the work done can be seen as the area under the PV curve for a given process, and we can also see that depending on the path taken to get from one point on the diagram to another the amount of work done will different.

First law for simple thermodynamic processes

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

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

ProcessConstant    ΔEintQW
Isothermal T 0 Q=W W=Q
Isobaric P Q-PΔV     ΔEint+PΔV     PΔV
Isovolumetric     V Q ΔEint 0
Adiabatic -W 0 -ΔEint

Free expansion

When we looked at the example of the gas expanding pushing the piston to derive the amount of work done in a volume change we considered work done by the gas against the piston. In the demo with the bottle we had to do work against the air.

If nothing opposes the expansion of the gas then no work is done by the gas when it expands. If the system is also insulated from heat then $Q=W=0$ and the internal energy $E_{int}$ does not change, meaning the temperature does not change, if the gas is ideal. Real gases do show a small decrease in temperature under free expansion, in real gases the internal energy depends a little on pressure and volume as well as temperature.

This situation is called free expansion.

Molar Specific Heat for Gases

We previously introduced the specific heat which gives the heat capacity heat capacity of a material per unit mass. Here we will use molar specific heats for gases at constant pressure $c_{P,m}$ and constant volume $c_{V,m}$ and explain the difference between these on the basis of the first law of thermodynamics.

If we increase the temperature of a gas by $\Delta T$ at constant volume $Q_{V}$ then the first law tells us that

$Q_{V}=\Delta E_{int}$

which means that all the heat applied to the system goes in to raising the temperature.

If we change the temperature the same amount constant pressure work will be done as the gas expands so

$Q_{P}=\Delta E_{int}+P\Delta V$

The temperature change is the same so $\Delta E_{int}$ is the same, so the difference between the heat in the two cases is then

$Q_{P}-Q_{V}=nc_{P,m}\Delta T-nc_{V,m}\Delta T=P\Delta V$

For an ideal gas $V=\frac{nRT}{P}$ means that $\Delta V=\frac{nR\Delta T}{P}$ so

$nc_{P,m}\Delta T-nc_{V,m}\Delta T=P(\frac{nR\Delta T}{P})$ → $c_{P,m}-c_{V,m}=R$

Let's recall that $R=8.314\,\mathrm{J/mol.K}$ and compare this to the table of specific heats on Wikipedia.

Constant volume molar heat capacity for an ideal gas

At constant volume all the heat added to the system goes in to raising the internal energy.

From the kinetic theory for an ideal monatomic gas the internal energy $E_{int}$ is given by

$E_{int}=\frac{3}{2}nRT$

so using the first law

$\Delta E_{int}=\frac{3}{2}nR\Delta T=nc_{V,m}\Delta T$ → $c_{V,m}=\frac{3}{2}R=12.471\,\mathrm{J/mol.K}$

We can also compare this to the table of specific heats on Wikipedia.

This prediction is very good for ideal monatomic gases, but gives values too low for more complicated molecules.

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Equipartition of energy

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.

We can see that diatomic molecules should thus have $c_{V,m}=\frac{5}{2}R=20.785\mathrm{J/mol.K}$

Based on this argument a bent molecule which has 3 degrees of rotational freedom might be expected to have $c_{V,m}=\frac{6}{2}R=24.942\mathrm{J/mol.K}$..but we can see they actually have a slightly higher specific heat than this.

Vibrational degrees of freedom

As well as moving and rotating molecules can vibrate. However these modes are typically “frozen out” in simple molecules at room temperature. As molecules get more complicated the vibrational modes start to contribute to the specific heat. If we cool a gas down we can the rotational degrees of freedom in gas can also become frozen. Quantum theory is required to explain why the number of active modes is dependent on temperature!

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Quasistatic adiabatic expansion of a Gas

The adiabatic expansion I demonstrated was far from quasistatic! However if we consider a quasistatic adiabatic expansion we can derive a useful relationship between pressure and volume in terms of the molar heat capacities we just considered. As $dQ=0$

$dE_{int}=dQ-dW=-dW=-PdV$

For any ideal gas $dE_{int}=nc_{V,m}dT$

So $nc_{V,m}dT+PdV=0$

If we take the ideal gas law $PV=nRT$ and differentiate both sides with respect to $T$

$P\frac{dV}{dT}+V\frac{dP}{dT}=nR$ → $P\,dV+V\,dP=nR\,dT$

Combining the two equations gives

$(c_{V,m}+R)P\,dV+c_{V,m}V\,dP=0$

$c_{P,m}P\,dV+c_{V,m}V\,dP=0$

which gives

$\frac{c_{P,m}}{c_{V,m}}\frac{dV}{V}+\frac{dP}{P}=0$

We define $\gamma=\frac{c_{P,m}}{c_{V,m}}$ and can integrate the previous equation to show that

$\ln P + \gamma \ln V = \mathrm{constant}$

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

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phy131studiof15/lectures/chapter21.1447856461.txt · Last modified: 2015/11/18 09:21 by mdawber
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