Physics 111 Lecture 39 (Walker: 17.6, 18.2) Latent Heat Internal Energy First Law of Thermodynamics May 8, 2009 Lecture 39 1/26 Latent Heats The heat required to convert from one phase to another is called the latent heat. The latent heat, L, is the heat that must be added to or removed from one kilogram of a substance to convert it from one phase to another. During the conversion process, the temperature of the system remains constant. Lecture 39 2/26
Latent Heat Heat of fusion, L F : heat (in J) required to change 1.0 kg of material from solid to liquid Heat of vaporization, L V : heat (in J) required to change 1.0 kg of material from liquid to vapor For melting a mass m of material that it already at the melting temperature, the heat required is Q = ml F For vaporizing a mass m of material that it already at the boiling point, the heat required is Q = ml V Latent Heats Lecture 39 4/26
Example How much heat must be removed from 2 kg of water at 0 C in order to freeze it? Q = ml F = (2 kg)(334 kj/kg) = 668 kj During the time the water is freezing, the temperature will stay at 0 C. Lecture 39 5/26 Phase Change & Energy Conservation Solving problems involving phase changes is similar to solving problems involving heat transfer, except that the latent heat must be included as well. Lecture 39 6/26
Latent Heat The total heat required for a phase change depends on the total mass and the latent heat: Problem Solving: Calorimetry 1. Is the system isolated? Are all significant sources of energy transfer known or calculable? 2. Apply conservation of energy. 3. If no phase changes occur, the heat transferred will depend on the mass, specific heat, and temperature change. Latent Heat 4. If there are, or may be, phase changes, terms that depend on the mass and the latent heat may also be present. Determine or estimate what phase the final system will be in. 5. Make sure that each term is in the right place and that all the temperature changes are positive. 6. There is only one final temperature when the system reaches equilibrium. 7. Solve.
Example 0.1 kg of ice at -10 C added to 1 kg of water at 30 C. Final temperature T f of system? (Assume water does not freeze.) Q lost-water = Q gained-ice m W c W (T Wi -T f ) = m I c I (10 C)+ m I L F + m I c W T f (1kg)(4.186 kj/kg- K)(30 C- T f ) = =(0.1kg)[(2.09kJ/kg- K)(10 C)+334 kj/kg +(4.186kJ/kg- K)T f ] T F = 19.7 C Lecture 39 9/26 Latent Heat On a molecular level, the heat added during a change of state does not go to increasing the kinetic energy of individual molecules, but rather to break the close bonds between them so the next phase can occur. In other words, the heat added during a phase change goes into internal potential energy.
Internal Energy U The internal energy U of an object is the sum total of all the random kinetic and potential energies of all the atoms/molecules in the object. The amount of internal energy depends on (a) the temperature, has more internal energy than (b) the number of molecules, and (c) the internal potential energy associated with interactions. Lecture 39 11/26 Internal Energy U The sum total of all the energy of all the molecules in a substance is its internal (or thermal) energy U. Temperature: measures molecules average random kinetic energy Internal energy U: total energy of all molecules (excluding external kinetic & potential energy) Heat Q: transfer of energy due to difference in temperature
Internal Energy An ideal gas has no internal potential energy since there are no interactions between molecules. The internal energy U is all kinetic, and for a monatomic gas, is : But since we know the average kinetic energy in terms of the temperature, we can write: where n is the number of moles of gas. Internal Energy - Molecular Gas If the gas is molecular rather than atomic, rotational and vibrational kinetic energy needs to be taken into account as well. Internal Rotational KE 1 N Iω 2 2 Internal Vibrational KE and PE
Internal Potential Energy The molecules in liquids and solids DO interact, so in those phases, there is internal potential energy. Water in the water vapor state at 100 C will have more internal potential energy than liquid water at the same temperature. The internal energy difference per kg is the Latent Heat of Vaporization (as we saw earlier). Likewise, water in the liquid state at 0 C will have more internal potential energy than ice at the same temperature. The internal energy difference per kg is the Latent Heat of Fusion. Lecture 39 15/26 Chapter 18 The Laws of Thermodynamics Lecture 39 16/26
First Law of Thermodynamics The first law of thermodynamics is a statement of the conservation of energy that includes energy transferred as heat (Q). If a system s volume is constant, and heat is added, its internal energy increases. Lecture 39 17/26 First Law of Thermodynamics If a system does work on the external world, and no heat is added, its internal energy decreases. Lecture 39 18/26
First Law of Thermodynamics Combining these gives the first law of thermodynamics. The change in a system s internal energy is related to the heat in, Q, and the work done, W, as follows: It is vital to keep track of the signs of Q and W. Lecture 39 19/26 The First Law of Thermodynamics The change in internal energy of a closed system will be equal to the energy added to the system minus the work done by the system on its surroundings. This is the law of conservation of energy, written in a form useful to systems involving heat transfer. Lecture 39 20/26
Example While 100J of heat is added to a cylinder containing gas, the gas does 80J of work by pushing up a piston. (a) Does the internal energy of the gas increase or decrease? How much? (b) Will the temperature of the gas increase? Lecture 39 21/26 Human Metabolism and the First Law If we apply the first law of thermodynamics to the human body: we know that the body can do work. If the internal energy is not to drop, there must be energy coming in. It isn t in the form of heat; the body loses heat rather than absorbing it. Rather, it is the chemical potential energy stored in foods. Lecture 39 22/26
Human Metabolism and the First Law The metabolic rate is the rate at which internal energy is transformed in the body. Lecture 39 23/26 Thermal Processes Thermal Processes are things we do to a system that involve thermal energy - adding heat, changing system temperature, etc. We will assume that all processes we discuss are quasi-static they are slow enough that the system is always in equilibrium. We also assume they are reversible: For a process to be reversible, it must be possible to return both the system and its surroundings to exactly the same states they were in before the process began. Lecture 39 24/26
Thermal Processes This is an idealized reversible process, with the system in a constant temp. heat bath. The gas is compressed; the temperature is constant, so heat leaves the gas. As the gas expands, it draws heat from the bath, returning the gas and the reservoir to their initial states. The piston is assumed frictionless. Lecture 39 25/26 Work Due to Volume Change Work is done by an expanding gas moving a piston a distance x. At constant pressure P, the force exerted on a piston of area A is PA, and the work is F x = P(A x), or in terms of volume change: Lecture 39 26/26
Work Due to Volume Change V positive (expansion): work done by gas on surroundings (W positive in 1 st Law) V negative (compression): work done on gas by surroundings (W negative in 1 st Law) Example -- What is work done by a gas when it expands by 1 liter (10-3 m 3 ) at atmospheric pressure (101 kpa)? Lecture 39 27/26 Constant Volume -- No Work If the volume stays constant, nothing moves and no work is done. Lecture 39 28/26
End of Lecture 39 For Monday, May 11, read Walker 18.3-4, 18.8 Homework Assignment 17c is due at 11:00 PM on Monday, May 11. Lecture 39 29/26