Lecture 4 The First Law of Thermodynamics Latent Heat Last lecture we saw that adding heat to an object does not always change the temperature of the object In some cases the heat causes a phase change instead example is ice melting solid liquid phase change for water while the ice melting the ice water mixture remains at 0oC heat flowing into the mixture is used to break bonds between molecules in the ice The amount of heat needed to complete the phase change depends on both the type and amount of material involved The latent heat of a material is defined as the amount of heat per unit mass needed to complete a phase change L Q m Of course there s usually more than one phase change for example H2O can go from solid ice to liquid water to gas steam The latent heat at the solid liquid transition is not the same as that at the liquid gas transition solid liquid is called latent heat of fusion Lf liquid gas is called latent heat of vaporization Lv Note the in the above the heat required to melt an object is exactly equal to the heat the object releases when it freezes Typically latent heat of vaporization is greater than latent heat of fusion The First Law of Thermodynamics Let s begin by reviewing some of the variables we can use to describe an object or a system of objects temperature volume mass pressure All such variables are called state variables i e each one tells us something about the state of the system There is another set of variables that describe the flow of energy into or out of the system we discussed one such transfer variable heat in the previous lecture Work The second transfer variable we ll study is work hopefully you remember this one from Physics 141 The new wrinkle we ll consider now is work done on systems that can change their shape including liquids and gases The definition of work is the integral of force applied over distance W F dr The tiny bit of work done when a force is applied over a tiny distance is dW F dr Work on a Gas Assume we have a cylinder filled with gas with a movable piston on top Piston y Gas x If the piston has area A the force on it is PA where P is the pressure in the gas Now imagine we push down slowly quasi statically on the piston this allows us to assume that the system remains in thermal equilibrium while we re pushing The work we do on the gas is dW F dr PAj dyj PAdy Since piston is moving at constant speed there s no net force on it That means we re applying a downward force exactly equal to the upward force from the gas Note that Ady is the tiny change in volume dV that results from our compression of the gas dW PdV We can integrate this to find the total work done Vf W PdV Vi We can t do the integral until we know how the pressure varies as we push on the gas in general it won t be constant Graphically we can find the work done from the area under the pressure vs volume curve P V Vf Vi The green area represents the work done in compressing the gas from Vi to Vf Note that you can t tell how much work was done just by looking at the initial and final pressures and volumes Consider these two cases where the initial and final pressures are the same Since T PV nR the initial and final temperatures are also the same P P V Pressure held constant during compression then increased Little work done V Pressure increased then held constant during compression Lots of work done So if all we have is the before and after pictures of the gas we can t tell how much mechanical work was done energy transfer by work depends on path followed on the P vs V graph Now consider energy transferred by heat into the gas Case 1 Initial state Very thin walls Environment outside the gas is also at Ti Piston and heat can flow easily through the thin walls Gas If we slowly lift the piston the gas remains in thermal equilibrium at Ti Ti enough heat flows in to make sure this is true Final state This leads to the following final state Piston Gas Now consider case 2 Ti Initial state Thin barrier Very thick walls Gas Ti No heat can flow through the thick wall If the thin barrier breaks the gas will expand to fill the entire volume Since there s no interaction with the outside world in this process the temperature remains constant This leads to the following final state The initial and final states are identical in cases 1 and 2 But in case 1 heat flows into the gas while in case 2 it does not Final state Gas Ti The First Law So we see that the energy transferred by work and the energy transferred by heat both depend on the process by which a gas goes from the initial to final state However we find that the sum of these two energy transfers does not depend on the process This allows us to define an additional state variable for the system the total energy stored inside Eint The internal energy can be changed both by mechanical work and heat and the first law of thermodynamics states that Eint Q W This is really a statement of conservation of energy in words it means that the change in energy inside a system equals the total energy flowing into or out of the system Cyclic Systems Consider a gas that takes the following path on the P vs V graph P V In this case the initial and final states are identical Since internal energy is a state variable this means that Eint 0 Using the First Law this tells us that Q W for this process in other words the energy transferred by work is equal and opposite to the energy transferred by heat