Recall

• For an ideal gas: PV = nRT
• Using a microscopic model of a gas, one finds that the pressure of the gas is related to the average kinetic energy of a molecule of gas: P = (2/3)(N/V)(½mv²_avg)

The Kinetic Theory of Gases (cont'd)

In this model, the temperature of a gas is related to the average kinetic energy of the molecules of the gas. To see this, rewrite the above expression as:
PV = (2/3)N(½mv²_avg).
Since the ideal gas law tells us that PV=nRT -- the same PV -- then we can equate nRT = (2/3)N(½mv²_avg). Use the relations n = N/N_A and R/N_A = k_B to get:
T = (2/3k_B)(½mv²_avg)
or, in the more standard form:
KE_avg = ½mv²_avg = (3/2)k_BT
This is another amazing result. It relates the average kinetic energy of a molecule of gas to the temperature of the gas as a whole!

Finally, a comment about the quantity v²_avg. This is the average of the square of the velocities. Often we want to speak of an average velocity (not squared). But the average velocity is zero, because there are as many molecules moving to the right (positive) as to the left (negative) on average, and they cancel to give an average velocity of zero. So instead we use the rms velocity. The letters rms stand for root-mean-square, which translates to the square root of the average of the square of the velocities. The rms velocity is
v_rms = Sqrt{v²_avg} = Sqrt{3k_BT/m} = Sqrt{3RT/M}
where M is the molar mass of the molecule.

Example: P10.38

A cylinder contains a mixture of helium and argon gas in equilibrium at a temperature of 150°C. (a) What is the average kinetic energy of each type of molecule? (b) What is the rms speed of each type of molecule?

(a) KE_avg = (3/2)k_BT = (3/2)(1.38x10^-23 J/K)(150+273°K) = 876x10^-23 J = 8.76x10^-21 J. Since the kinetic energy doesn't depend on the type of atom or molecule (i.e. no mass dependence), the result is the same for both gases.

(b) v_rms = Sqrt{3RT/M}
This does depend on the molecular mass. M_He = 4.0x10^-3 kg/mol and M_Ar = 39.948x10^-3 kg/mol. The resulting rms speeds are:
v_{rms He} = Sqrt{3(8.31 J/mol K)(423K)/(4.0x10^-3 kg/mol) = 1620 m/s
v_{rms Ar} = Sqrt{3(8.31 J/mol K)(423K)/(40.x10^-3 kg/mol) = 513 m/s.

Chapter 11: Heat

The Mechanical Equivalent of Heat

As alluded to earlier, heat is a form of energy. Carefully constructed experiments have shown that when mechanical energy is "lost" due to non-conservative forces, that it turns into heat. By expanding our concept of energy to include heat -- along with kinetic energy of translation and rotation, gravitational potential energy, and elastic potential energy -- we regain conservation of energy in a more general form.

Heat energy is (still) commonly measured in calories. One calorie is the energy needed to raise the temperature of 1 gram of water from 14.5°C to 15.5°C. To convert to joules, the SI unit of energy, 1 cal = 4.186 J. The Calories (big c) used for food equal 1000 calories (little c).

Specific Heat

If we add heat to something, generally its temperature increases. The specific heat is a measure of the relation between heat energy and temperature. Since the amount of heat added must also depend on how much of the substance there is, we also include the mass in the relation:
c = Q/mDT.
Table 11.1 contains a list of specific heats.

Conservation of Energy: Calorimetry

First, we want to consider problems where heat energy is transferred between objects. A controlled system, that is allows energy to be transferred between well defined objects, is called a calorimeter. We will consider cases where heat leaves one substance and goes to a second. What is being transferred is heat energy, Q = mcDT. The temperatures, masses and specific heats of the substances may be different, but the heat that leaves on substance got to another:
Q_out = Q_in
For now, make both energies positive, that is, make DT positive. The following examples illustrate the general procedure.

Example: P11.11

Lead pellets, each of mass 1.00 g, are heated to 200°C. How many pellets must be added to 500 g of water that is initially at 20.0°C to make the equilibrium temperature 25.0°C? Neglect any heat transfer to or from the container.

The amount of heat lost by the lead pellets when cooled to 25°C must equal the heat needed to raise the temperature of water to 25°C. First, how much heat must be added to the water?
Q_w = c_wm_w(T_f - T_w)
where:

• c_w = specific heat of water = 4186 J/kg°C
• m_w = mass of water = 0.50 kg
• T_f = final temperature = 25°C
• T_w = initial water temperature = 20°C
• T_Pb = initial lead temperature = 200°C
• c_Pb = specific heat of lead = 128 J/kg°C
• m_p = mass of a lead pellet = 0.001 kg
• N = number of lead pellets needed.

The heat lost by N lead pellets is:
Q_Pb = Nm_pc_Pb(T_Pb - T_f)
Set the two heats equal to each other and solve for N:
N = (c_w/c_Pb)(m_w/m_p){(T_f - T_w)/(T_Pb - T_f) = (4186/128)(0.50/0.001){(25-20)/(200-25)} = 467.

Example: P11.16

A 100 g aluminum calorimeter contains 250 g of water. The two are in thermal equilibrium at 10°C. Two metallic blocks are placed in the water. One is a 50 g piece of copper at 80°C. The other sample has a mass of 70 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20°C. Determine the specific heat of the unknown second sample.

The heat lost by the metallic blocks must equal the heat gained by the water and aluminum calorimeter.
Q_Al+w = m_Alc_Al(T_f - T_Al) + m_wc_w(T_f - T_w) = (m_Alc_Al + m_wc_w)(T_f - T_Al+w).
Q_blocks = m_Cuc_Cu(T_Cu - T_f) + m_xc_x(T_x - T_f)
Again, equate the two heats and this time solve for c_x, the specific heat of the unknown sample. All other quantities are known.
c_x = {(m_Alc_Al + m_wc_w)(T_f - T_Al+w) - m_Cuc_Cu(T_Cu - T_f)}/(m_x(T_x - T_f)) = {((0.10kg)(900J/kg°C) + (0.25kg)(4186J/kg°C))(20-10°C) - (0.05kg)(387J/kg°C)(80-20°C)}/(0.07kg)(100-20°C) =1800 J/kg°C.

Latent Heat and Phase Changes

When a substance undergoes a change of phase, heat is absorbed or given off without a change of temperature. Familiar phase changes are from solid to liquid (ice melting to water), the reverse of liquid to solid (water freezing to ice), liquid to gas (water boiling away to steam) or its reverse of gas to liquid (steam condensing to water). The more rare phase change from solid to gas is possible, as are a number of more specialized cases such as:

• from (ferro)magnetic to non-magnetic (paramagnetic)
• from superconducting to normal

Phase changes generally occur because the internal structure of a substance is changed. We characterize the energy absorbed of given off by the latent heat, L, so that Q = mL is the amount of heat for a mass m of the substance. When the phase change is between solid and liquid, use the latent heat of fusion, L_f, and when the change is between liquid and gas, use the latent heat of vaporization, L_v. Table 11.2 is a list of latent heats for a variety of substances.

If in a calorimetry problem one of the substances undergoes a phase change, then the calculation of the heat lost of gained must be modified to include the energy involved in the phase change.

Example: P11.20

A 50 g ice cube at 0°C is heated until 45 g has become water at 100°C and 5.0 g has become steam at 100°C. How much heat was added to accomplish this?

First, what are the steps required to go from the ice cube initial state to the water plus steam final state?

1. Melt the ice into water
2. Heat the water from 0°C to 100°C
3. Turn 5.0g of the 100°C water into steam.

1. The heat required to melt 50.0g of ice is Q₁ = mL_f = (0.05kg)(3.33x10⁵J/kg) = 1.7x10⁴J.
2. The heat required to raise the temperature of 50g of water from 0°C to 100°C is Q₂ = mcDT = (0.05kg)(4186J/kg°C)(100°C) = 21000 J = 2.1x10⁴J.
3. The heat required to vaporize 5.0g of water is Q₃ = mL_v = (0.005kg)(2.26x10⁶) = 1.1x10⁴J.

The total heat required is Q_tot = Q₁ + Q₂ + Q₃ = (1.7 + 2.1 + 1.1)x10⁴J = 4.9x10⁴J.

Heat Transfer by Conduction, Convection, and Radiation

There are three ways to transfer heat: conduction, convection, and radiation.

• Conduction occurs when heat propagates from a source to a destination through an intervening material. For instance, water is heated by placing it in a container and placing the container of a stove. Heat is transferred through the pot to the water.
• In convection, heat is transferred by currents in a gas or liquid. For example, bread is baked in an oven by the heated air in the oven. In a traditional oven, the convection is natural; in a convection oven, the air is forced to circulate by a fan, and this is called forced convection.
• Radiative heat transfer occurs through the emission and absorption of electro-magnetic radiation, in particular, infrared radiation. This is the manner in which food is broiled or grilled. A microwave oven also uses radiation, but instead of infrared, microwave radiation is used.

Conduction

In conduction, an amount of energy, Q, is transported through a material in a given amount of time Dt. The heat transfer rate, H, is defined as H=Q/Dt.

Now imagine heat passing through a block of material with area A and thickness L, one side of the block at temperature T₁ and the other side at temperature T₂. It is found that the heat transfer rate is proportional to the area and the temperature difference, and inversely proportional to the thickness.
H = Q/Dt = kA(T₂ - T₁)/L
where k is a constant for the material called the thermal conductivity. Table 11.3 lists values of thermal conductivity, k, for a number of materials.

Convection

Radiation

All objects at a finite temperature emit energy as electro-magnetic radiation. Light is radiation with wavelengths in the visible band; most heat transfer occurs through radiation in the infrared band. The heating element of a toaster glows red, and light bulbs are hot.

The rate at which an object emits radiation is given by Stefan's law:
P = sAeT⁴
where

• P is the power radiated in Watts,
• s is a constant equal to 6.0x10^-8W/m²K⁴,
• A is the surface of the object in square meters,
• e is the emissivity, a material dependent constant between 0 and 1, and
• T is the object's temperature in kelvins

When an object is surrounded by other objects at finite temperature, then it not only emits radiation, but will also absorb radiation from the surroundings. The net rate at which energy is gained or lost is the difference between the rate of emission, and the rate of absorption:
P_net = sAe(T⁴-T₀⁴)
where T₀ is the temperature of the surroundings.

Example: P11.37

A sphere that is to be considered as a perfect black-body radiator has a radius of 0.060m and is at 200°C in a room where the temperature is 22°C. Calculate the net rate at which the sphere radiates energy.

A blackbody is an ideal absorber of radiation and has an emissivity of 1.0. Use P_net = sAe(T⁴-T₀⁴). The area of a sphere is 4pr², therefore:
P_net = (6.0x10^-8W/m²K⁴)4p(0.060 m)²(1.0)((473°K)⁴ - (295°K)⁴) = 110 W.