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Chapters 17, thermodynamics
Lecture 24 Goals: Chapters 17, thermodynamics Assignment HW-10 due Tuesday, Nov 30 Monday: Read Chapter 18 Third test on Thursday, December 2 1
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First Law of Thermodynamics
ΔU+ΔK+ΔEthermal=ΔEsystem=Wexternal+Q For systems where there is no change in mechanical energy: ΔEthermal =Wexternal+Q Last time, we discussed the first law of thermodynamics. We argued that, there are two ways that you can change the total energy of a closed system. You can either do work on the system, or you can exchange heat, which is the thermal energy transfer. The total energy of the system is its potential energy+kinetic energy+thermal energy, which is proportional to the temperature. For thermodynamic discussions, we concentrate on systems where there is no change in mechanical energy. This equation therefore reads, the change in thermal energy of a system is the work that you do + the heat that you provide. Heat and work both has units of energy, in SI units, Joules, but historically, a different unit calories has been used for heat. 1 calorie=4.186 Joules 1 food calorie=1000 calories
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Isochoric process (const V)
Pi W=0 Work is: -(the area under the curve) Pressure Using ideal gas law: PV=nRT Tf < Ti Pf Volume From first law of thermodynamics we conclude: ΔEthermal =W+Q<0 Q < 0 Now, we are going to go through several processes in detail to make sure that we understand all the concepts that go into the first law of thermodynamics. Let’s first consider an isochoric process, a constant volume process. In this particular example, the volume is fixed, and the pressure of the gas drops from Pi to Pf. We are also considering a sealed container, the amount of gas does not change, the number of moles does not change. Now, first observation is that the work done on the gas, is the area under the curve, which as we discussed before, is zero. So there is no work done on the gas. Now the second observation is that, from the ideal gas law, PV=nRT. When we compare these two points, the volume does not change, n does not change, R is a constant, so since Pf<Pi, we must conclude, Tf<Ti. So the thermal energy of the gas decreases. 5) From first law of thermodynamics, the change in thermal energy is W+Q, since W is zero, we conclude that Q must be less than 0. So during this process, heat was transferred from the gas to the environment
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Isothermal expansion (const T)
Pi ΔEthermal=0 Temperature does not change Pressure Pf W<0 Work done on the gas is: Vi Vf Volume From first law of thermodynamics we conclude: ΔEthermal =W+Q=0 Q=-W Now, let’s consider another example. Here we will focus on an isothermal expansion process, where the temperature does not change. So we start at an initial point Pi Vi, and the gas expands to a final point Pf Vf. Now the first observation is that because there is no change in temperature, the thermal energy of the gas does not change. Now, because the gas is expanding, the work done on the gas is negative. When we integrate this curve from Vi to Vf, it is just the area under this curve, so it is a positive number, and the work is – the area, so we conclude that Work done on the gas is a negative number Now, finally, from first law of thermodynamics, we conclude that Q>0. Which means that there was heat energy transferred from the environment to the gas, so we need access to thermal energy for this to happen. Note that heat energy transfer does not necessarily result in a change in temperature.
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Work in a isothermal process
Pi Pressure Pf Vi Vf Volume So we argued that the work done on the gas would be a negative quantity, we can actually find its exact value by just evaluating the area under the isothermal curve. In this slide, I would like to go through this argument. Work is – the area under the curve, PdV Now, from ideal gas law, P=nRT/V. Now, since this is an isothermal process, T is a constant, so we can take these out of the integral.
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Isothermal compression
ΔEthermal=0 ΔEthermal <0 ΔEthermal >0 What is the change in thermal energy? Pf Pressure Pi Vf Vi Volume What is the work done on the gas? A) W=0 B) W<0 B) W>0 What direction is the thermal energy flow? A) Q=0 B) Q<0 B) Q>0
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Adiabatic processes Q=0
Processes where there is no thermal energy transfer to or from the system: This does not mean that there is no change in temperature. For adiabatic processes: Q=0
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Thermal Properties of Matter
Now, so far we have been focusing exclusively on gases, and so far we have only considered cases where the gas remains an ideal gas, and the change in thermal energy simply means changing the temperature of the gas. However, when you change the thermal energy of a system, there can also be a phase transition happening. So we would like to thermal properties of matter a little bit more closely. Suppose that you get a piece of ice and you start adding energy to it. Now if you measure the temperature of the ice as a function of energy added, you observe a graph that looks like this. Let’s assume that we start the experiment with ice at -30 degrees celcius. First you increase the temperature of the ice. But once you hit zero degrees celcius, you will start converting ice into water, so this is the point where the phase transition occurs. And then once you convert all the ice to water, you will start increasing the temperature of the water, and so on. Now, the thermal energy of the system is increasing throughout the whole graph, we are adding energy to the system. But what happens is that, in these linear regions, the enery is used to increase the temperature, average motion of the molecules. In these flat regions, the energy is used not to increase the speed of the molecules, but instead to break molecular bonds, so that the substance enters a new phase.
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Heat of Transformation
Latent heat of transformation L is the energy required for 1 kg of substance to undergo a phase change. (J / kg) Q = ±ML Heat of vaporization for water is L=2.3x106 J/kg. Q = +2.3x106 J 1) In the flat regions of the graph, we ask the question, how much heat is required to cause a phase change, to for example, melt 1 kg of ice, or to evaporate 1 kg of water. 1 kg Q = -2.3x106 J
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Temperature change and specific heat
Specific heat c of a substance is the energy required to raise the temperature of 1 kg by 1 K. (Units: J / K kg ) Q = M c ΔT Specific heat for water is c=4190 J/ K kg. Q = J 1 kg T+1 K 1 kg T K Q = J
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Exercise The specific heat (Q = M c ΔT) of aluminum is about twice that of iron. Consider two blocks of equal mass, one made of aluminum and the other one made of iron, initially in thermal equilibrium. Heat is added to each block at the same constant rate until it reaches a temperature of 500 K. Which of the following statements is true? (a) The iron takes less time than the aluminum to reach 500 K (b) The aluminum takes less time than the iron to reach 500 K (c) The two blocks take the same amount of time to reach 500 K
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Specific heat for gases
For gases we typically use molar specific heat (Units: J / K mol ) Q = n C ΔT For gases there is an additional complication. Since we can also change the temperature by doing work, the specific heat depends on the path. Q = n CV ΔT (temperature change at constant V) Q = n CP ΔT (temperature change at constant P)
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Relationship between CP and CV
For path 1 Pressure ΔEthermal =W+Q=nCVΔT 2 P For path 2 Ti 1 ΔEthermal =W+Q =-PΔV+nCPΔT =-nRΔT+nCPΔT Tf ΔV Because the initial and final temperatures are the same, the change in thermal energy will be the same for the two processes Once again, the change in the thermal energy is the same for both paths, but the heat required for the two paths is different simply because you are doing work in one path and you are not in the other one. Volume Equating both paths we get CP=CV+R
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Monoatomic vs diatomic gases
Molar specific heats of all monatomic gases are around CV≈12.5 J/K mol Molar specific heats of all diatomic gases are around CV≈20.8 J/K mol
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Energy transfer mechanisms
Thermal conduction (or conduction) Convection Thermal Radiation 1) Although we have talked a lot about heat energy transfer, we have not discussed much the mechanisms of energy transfer. So when I put my hand in hot water, there is heat energy transferred to my hand. Also, when I go out in the sun, I feel warmer, and there is again heat energy transfer. So what are the mechanisms for heat energy transfer.
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Thermal conduction Energy transferred by direct contact.
Rate of energy transfer ( J / s or W ) Through a slab of area A and thickness Dx, with opposite faces at different temperatures, Tc and Th Q / t = k A (Th - Tc ) / x k :Thermal conductivity (J / s m °C) 1) Thermal conduction: this is what happens when I put my hand in warm water.
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Energy transfer mechanisms
Convection: Energy is transferred by flow of substance Natural convection: from differences in density Forced convection: from pump of fan Radiation: Energy is transferred by photons e.g.: infrared lamps Stefan’s Law s = 5.710-8 W/m2 K4 , T is in Kelvin, and A is the surface area e is a constant called the emissivity P = A e T4 (power radiated)
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