There are three ways to transfer heat: conduction, convection, and radiation.
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.
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
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:
Pnet = sAe(T4-T04)
where T0 is the temperature of the surroundings.
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 Pnet = sAe(T4-T04).
The area of a sphere is 4pr2, therefore:
Pnet = (6.0x10-8W/m2K4)4p(0.060 m)2(1.0)((473°K)4 - (295°K)4) = 110 W.
The first law of thermodynamics concerns the internal energy of a system, and the ways that energy can be changed. First, the internal energy of something is the energy contained within its atoms and molecules. This does NOT include the kinetic energy or potential energy of the object or substance itself. We are now thinking about the stored in the constituent atoms and molecules of a "system".
For example, 100g of ice and 100g of hot water sitting on a table have the same gravitational potential energy, mgh, but they have different temperatures, and different amounts of internal energy. The H2O molecules in the ice have less energy on average than the molecules in the hot water.
The internal energy of a system can be changed by:
If U stands for the internal energy, Q for the heat added to the system, and W for the work done BY the system, then the first law is expressed by:
DU = Uf - Ui = Q - W.
If heat is added to the system, the internal energy increases.
If the system does work, it must lose a corresponding amount of internal energy.
The only form of energy possessed by "molecules" of a monatomic ideal gas is translational kinetic energy. Using the results from the discussion of kinetic theory in section 10.6, show that the internal energy of a monatomic ideal gas at pressure P and occupying volume V may be written as U = (3/2)PV.
First, examples of monatomic gases are helium, neon, and argon. Monatomic gases are composed of single atoms -- the atoms do not bind together into molecules. So the phrase "molecules of a monatomic ideal gas" is a bit awkward from my point of view. For our purposes, you can replace that phrase with "atoms of a noble gas".
Again, for our purposes, the atoms of a noble gas act like point masses.
They can move about, bounce off the container and each other, but there's nothing to rotate, and no internal vibrations in which to store energy.
Thus, all the internal energy is translational kinetic energy, ½mv2avg.
And in chapter 10 we learned that the average kinetic energy of an atom in a gas is KEavg = ½mv2avg = (3/2)kBT.
The internal energy of a gas is the sum of the energy of all the constituents:
U = SKEi = N KEavg = nNAKEavg = (3/2)n(NAkB)T.
Recall that kB = R/NA, so the quantity in parenthesis is R.
Making use of the ideal gas law, PV = nRT, we find:
U = (3/2)nRT = (3/2)PV.
This result is useful in several of the homework problems.
A gas at pressure P, changing volume by an amount DV, does work W=PDV. If we make a plot with volume on the horizontal axis and pressure on the vertical axis, then we see that the work done by the gas in an expansion is equal to the area under the curve from the point at Pi, Vi to Pf, Vf. Such a plot is called a PV diagram.
If the pressure is not constant while the volume changes, the work can still be calculated as the area under the PV curve. To do this we must know the shape of the PV curve, that is, we must know how the pressure and volume changed in going from the initial point to the final point.
Sketch a PV diagram of the following process. (a) A gas expands at constant pressure P1 from volume V1 to volume V2. It is then kept at constant volume while the pressure is reduced to P2. (b) A gas is reduced in pressure from P1 to P2 while its volume is held constant at V1. It is then expanded at constant pressure P2 to a final volume, V2. (c) In which of the processes is more work done? Why?
Sketch (a) and (b). (c) Since P1>P2, the area under the PV curve for (b) is less than (a), and the work done in process (b) is less than (a).
A gas expands from I to F along the three paths indicated in the figure. Calculate the work done by the gas along paths (a) IAF, (b) IF, and (c) IBF.
In each case, the work done by the gas is equal to the area under the PV curve. To get the work in joules, the pressure must be expressed in pascals = N/m² and the volume in m³. The conversion factors are 1 atm = 1.01x105Pa, and 1 liter = 10-3m³. Therefore Pi = 4.04x105Pa, Pf = 1.01x105Pa, Vi = 2.0x10-3m³, and Vf = 4.0x10-3m³.
(a) The area under the curve IAF is Wa = Pi(Vf - Vi) = 808 J.
(b) Recall that the area of a triangle is ½base x height. The area under the curve IF is Wb = ½(Pf - Pi)(Vf - Vi) + Pf(Vf - Vi) = ½(3.03x105Pa)(2.0x10-3m³) + (1.01x105Pa)(2.0x10-3m³) = 505 J.
(c) The area under the curve IBF is Wc = Pf(Vf - Vi) = 202 J.
A gas is enclosed in a container fitted with a piston of cross-sectional area 0.150 m². The pressure of the gas is maintained at 6000 Pa as the piston moves inward 20.0 cm. (a) Calculate the work done by the gas. (b) If the internal energy of the gas decreases by 8.00 J, find the amount of heat removed from the system during the compression.
(a) Since the volume of the gas decreases, work is done on the gas, or the work done by the gas is negative. The amount of work done is W = PDV = PADs = (6000 Pa)(0.150 m²)(-0.200 m) = -180 J.
(b) While the gas is compressed, heat is added or removed such that the pressure remains constant. Using the first law, DU = Q - W, or Q = DU + W = -8.00J + (-180 J) = -188 J. Remember that Q is defined as the heat added to the system, a negative Q implying that heat is removed. Therefore the amount of heat removed is 188 J.
One of the earliest and most important applications of thermodynamics is the understanding of engines. For our purposes, a heat engine is a system that goes through changes in pressure, volume, and temperature, in a cyclic fashion. That is, the system repeatedly follows the same changes in P, V, and T, returning to its starting conditions to begin another cycle.
By far the most common engines involve a system consisting of a gas, so this is what we will concentrate on.
We can represent the state of the gas on a PV diagram.
(Since, for a given quantity of an ideal gas, T = PV/nR, specifying PV fixes T.)
A cyclic process traces out a closed curve on the PV diagram. (diagram)
The net work done by the engine per cycle equals the area enclosed by the curve,
Wcycle = Area enclosed in PV curve.
A heat engine can be represented schematically as a system connected to a hot reservoir and a cold reservoir. The system absorbs an amount of heat Qh from the hot reservoir, gives off an amount of heat Qc to the cold reservoir, and produces an amount of work W on each cycle. Since the thermodynamic process of the engine is cyclic, it returns to its starting condition after each cycle. Being in the same condition means that the internal energy of the system is the same, or DUcycle = 0 for a complete cycle.
The first law tells us that DUcycle = Qcycle - Wcycle, or, since DUcycle = 0, Qcycle = Qh - Qc = Wcycle. This is just conservation of energy. The system takes in an amount of heat energy Qh, and gives off an amount of heat energy Qc and work Wcycle.
The efficiency, e, of the engine is defined as the fraction of the heat energy taken in that is converted to work:
e = W/Qh = (Qh - Qc)/Qh = 1 - Qh/Qc.
If the engine converts all the input energy to heat (W = Qh) then the efficiency is 100%.
If no work is produced then the efficiency is 0.
As applied to heat engines, the second law of thermodynamics says that the amount of work produced by a heat engine is always less than the amount of input energy, or, it is impossible for an engine to be 100% efficient. Mathematically, W < Qh, never equal to.