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Lecture17 - Heat and The First Law of Thermodynamics...

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Unformatted text preview: Heat and The First Law of Thermodynamics Heat and mechanical energy Un<l about the early 1800s, nobody thought of connec<on between heat energy and mechanical energy. They were thought to be different quan<<es. Heat was thought of a kind of invisible fluid. Mechanical energy was measured in ergs while heat was measured in calories. However with the works of Benjamin Thomson and more importantly later by James Joule, it was realized their equivalence and mechanical work produce heat. Joule, with his famous experiment named aNer him established the factor for the equivalence, which we now write as: 1 calorie = 4.186 joules That is, what used to be 1 calorie of heat energy is now equivalent of 4.186 joules of mechanical energy. Chapter 20 Heat and specific heat capacity Heat is energy transferred between two systems because of a temperature difference exis<ng between the systems. Heat transfer to and from a system results in temperature difference, except in the case of phase change of the system. The quan<ty of energy required to raise the temperature of a given mass of a substance by some amount varies from substance to substance. We need a quan<ty that expresses this fact numerically – that is it is the property of a substance used. In fact we have two quan<<es: Heat capacity and specific heat capacity. The former depends on the mass of the substance while the later is defined to be mass independent (by adding an extra qualifier ‘per unit mass’ as seen the defini<on below). Heat capacity C of a substance is defined as the amount of energy needed to raise its temperature by 1oC. Thus the amount of energy Q required to raise the temperature of a substance of heat capacity C by ΔT is given by Q = C ΔT = C (T f − Ti ) C is expressed in J/K or J/oC. Heat and specific heat capacity (contd…) Specific heat capacity of a substance is defined as the amount of energy required to raise a temperature by 1oC of 1 kg sample of the substance. In other words, specific heat capacity is the heat capacity per unit mass. Specific heat capacity is denoted by c and is expressed in J/(kg.K) or J/(kg.oC). Examples: for water, c = 4186 J/(kg.oC) for iron, c = 448 J/(kg.oC) Thus, water requires more energy (more than nine <mes) to raise the temperature by the same amount if we compare for the same mass. From the defini<on of specific heat capacity, it follows that the heat required to raise the temperature of a substance of mass m and specific heat capacity c by ΔT is given by: Q = mcΔT = mc(T f − Ti ) Example 1 How much energy is required to raise the temperature of a 500g iron ball to 100 oC if it is at 20 oC ini<ally? Q = mc(T f − Ti ) = (0.5 kg )( 448 Jkg −1oC −1 )(100 o C − 20 o C ) = 17920 J ≈ 1.8 kJ Example 2 A cowboy fires a silver bullet with a muzzle speed of 200m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet and doesn’t flow into the wall. What is change in temperature of the bullet aNer it hits the wall? Here, the bullet gains internal energy (heat) at the cost of all of its KE. So if we include internal energy energy, the conserva<on of energy means: ΔK + ΔEint = 0 , ΔEint = Q = mcΔT 12 1 mv ) + mcΔT = 0 ⇒ mv 2 = mcΔT 2 2 2 2 v ( 200 m / s) ⇒ ΔT = = = 85.5 oC o 2c 2( 234 J C / kg ) ⇒ (0 − Latent heat Ice melts into water at 0oC . Water evaporates at 100oC . These are example of phase change. Water changes from solid phase (ice) to liquid phase (water), in the former example. Phase change, not just a temperature change, requires energy. Therefore, if energy is spent on changing the phase of the substance, temperature of the substance does not change even if it is absorbing heat from the surrounding. For example, once the water is heated to 100oC, the temperature cannot be increased from 100oC, no maeer how much extra energy we supply to it. The energy is spent on dissocia<ng the water molecules thus on conver<ng the water to water vapor. If the vapor is confined and if all water is converted to vapor, then the extra heat supplied goes to increase the temperature of water vapor. We have the same situa<on in mel<ng ice. The absorp<on of heat from the surroundings does not result in temperature change un<l all ice melts. Latent heat (contd…) The heat required for the change of phase at a temperature is called the latent heat of the substance for that phase change. Thus we have latent heat of fusion of ice, latent heat of evapora<on of water. Heat required to change the phase is given by where L is the latent heat. It is expressed in J/kg. Latent heat of fusion of ice, Lf = 3.33 × 105 J/kg. Latent heat of vaporiza<on of water, Lv = 2.26 × 106 J/kg Q = mL Example How much heat is required to convert 200g of ice at  ­20oC to 200f of steam at 120oC ? Calorimetry If a small hot iron ball is dropped into a pail of water, the ball loses energy while the water gains energy. Net heat transfer stops when the system comes in thermal equilibrium. We can find the specific heat capacity of iron from the measured values of ini<al and final temperatures and from the specific heat capacity of water. This kind of measurements is called the calorimetry and the device in which heat transfer occurs (here the pail) is called calorimeter. The basic principle of calorimetry is the principle of conserva<on of energy. Here it is wrieen expressed as: Qcold = −Qhot which means that the heat energy leaving the hot part of the system is equal to the that entering in the cold part. Sign conven<on:  ­ ve: leaving + ve: entering Example A 0.050 kg piece of a metal is heated to 200.0 oC and then dropped into a beaker containing 0.400 kg of water ini<ally at 20.0 oC. If the final equilibrium temperature of the mixed system is 22.4 oC, find the specific heat of the metal. Heat and work Heat is not only a way of energy transfer between a system and the surroundings. Energy in can be transferred by doing work as well. Consider a simple system: a gas in equilibrium under pressure p in a volume V. Assume that the gas is in a container fieed with a movable Piston. Now suppose that the gas expands slowly (slow enough to approximately maintain thermal equilibrium in the process) pushing the piston through a distance dy. The gas expands from V to V + Ady, where A is the cross ­sec<on area of the piston. Work done by the gas on the piston is: dW = F ⋅ ds = ( pA)(dy) = p( Ady) = p dV Total work done by the gas in the volume change from Vi to Vf is then: Vf W= Vi ∫ p dV Sign conven3on: work done by the system (gas here): +VE work done on the system:  ­VE since dV is +ve in expansion and nega<ve in compression. PV  ­ diagrams W= Vf Vi ∫ p dV In order to calculate W, we see that we need to know ini<al and final volume and the a func<on p(V). If we know the func<on, we can also plot p vs V. The resul<ng diagram is called a pV ­diagram. Then it is clear that the work done equals the area under the pV curve between Vi and Vf. However, we can experimentally get to the final state f from the ini3al state i in some other ways. The diagrams below shows an example. What does this mean? This means that the work done in taking the gas from ini<al state to the final state depends on the path chosen to get there. Thus the work done by a system depends on the ini<al state, the final state and on the path followed by the system between these states. It follows that work done is not the property of the system (like temperature and volumes). It can also be seen that heat transfer also not a property of a system – it depends on the process chosen between the states. First law of thermodynamics Suppose that a system (for example, a gas in a cylinder with piston) undergoes a change from an ini<al state to a final state. During the change, energy transfer by heat Q to the system occurs and work W is done by the system. It is found that the quan<ty Q – W depends on the ini<al and final states only and not on the paths connec<ng these states. In other work Q – W is property of the system. The quan<ty Q – W is called the change in internal energy of the system ΔEint. Thus we have ΔEint = Q  ­ W ...
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