Review Questions Why can a person remove a piece of dry aluminum foil from a hot oven with bare fingers without getting burned, yet will be burned doing so if the foil is wet. Equal quantities of alcohol and water at different temperatures are mixed together and the mixture is allowed to come to equilibrium. Which statement(s) is/are true? 1.The alcohol will have the larger temperature change. 2.The water will have the larger temperature change. 3.The water and the alcohol will each have the same temperature change.
Phase Changes and Latent Heat When a material changes from a solid to a liquid, or from a liquid to a gas, a certain amount of energy is involved in the change of phase. This energy must be supplied or removed from the system to cause the molecular rearrangement, and is called the latent heat. Phase transitions occur at constant temperature at the melting or boiling point.
Phase Changes SOLID LIQUID GAS L F L latent heat of fusion latent heat of vaporization Phase transitions occur at a fixed, constant temperature until they are complete. Heat supplied to or removed from a system during a phase transition causes no temperature change. For example, at sea level: ice melts in liquid water at 0 ºC water boils at 100 ºC
Heat and Phase Transitions The heat Q that must be supplied to or removed from a system to cause a mass m of material to undergo a phase change: Q = ml m = mass of material (kg) L = latent heat of material (J/kg)
Latent Heats and Phase Transition Temperatures of Some Substances (at atmospheric pressure) Substance T melting ( C) L F (J/kg) T boiling ( C) L (J/kg) Nitrogen -210 0.26 x 10 5-195.8 2.00 x 10 5 Water 0 3.33 x 10 5 100 22.6 x 10 5 Silver 961 0.88 x 10 5 2193 23 x 10 5
Temperature changes and heat needed to convert 1 kg of ice at -40ºC to steam above 100ºC
Calorimetry Example with Phase Change What mass of steam initially at 130 C is needed to warm 200 g of water in a 100-g glass container from 20.0 C to 50.0 C? Latent heat of vaporization of water: L = 2.26 x 10 6 J/kg Specific heat of water: c W = 4186 J/(kg C ) Specific heat of steam: c S = 2010 J/(kg C ) Specific heat of glass: c G = 840 J/(kg C )
Review Questions Is it possible for the temperature of an object to remain constant even though heat flows into it or away from it? Citrus groves are protected from freezing during cold spell by spraying the trees with water. Why is this effective?
First Law of Thermodynamics Some Definitions: System: an object or set of objects. Closed System: no mass enters or leaves. Open System: mass may enter or leave. Isolated system: no energy enters or leaves. First Law of Thermodynamics: U = Q - W change in heat transfer work done internal energy
Sign Conventions for First Law of Thermodynamics Q is + if heat is added to system. Q is if heat is lost by system. W is + if work is done by the system. W is if work is done on the system.
Classes of Thermodynamic Processes ISOTHERMAL PROCESS: T = 0 ADIABATIC PROCESS: Q = 0 ISOBARIC PROCESS: P = 0 ISOCHORIC PROCESS: = 0 (ISOOLUMETRIC)
Work Done by Gases e.g. piston in a cylinder moves by dl due to expansion of gas at pressure P Force on Piston: F = P A Incremental Work Done: dw = F dl = P A dl = P d Total Work Done : W = dw = f i Pd Units for P : Pa m 3 = (N/m 2 ) m 3 = N m = J
Work Done by a Gas Work done depends on the path taken. W = B Pd = area under P - A curve
Work Done by a Gas Depends on the Path Taken in P- Space Work done along path ADB (isochoric AD plus isobaric DB) is less than that done along isothermal path AB.
Example: work done by a gas: Problem 19-38 Consider the two-step process shown. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.5 atm. Then the gas expands at constant pressure, from a volume of 6.8 L to 10.0 L, where the temperature reaches its original value. Calculate: The total work done by the gas in the process. The change in internal energy of the gas. The total heat flow into or out of the gas.
Work Done by an Ideal Gas in an Isothermal Process For an isothermal process: T = 0 (T is constant) W = B Pd P = nrt P = A nrt W B d B = nrt = nrt A A d = nrt ln B A
Molar Specific Heats of Gases To raise the temperature of n moles of a gas by T requires heat Q: Q = n C T (at constant volume) Q = n C P T (at constant pressure) C P >C for all gases [units: cal. mole -1 K -1 )]
Molar Specific Heat of an Ideal Gas at Constant olume For an ideal gas, U = (3/2)n R T If the temperature T is increased by T U = (3/2)n R T If is constant, W=0 U = Q W = Q Q = n C T = (3/2)n R T C = (3/2) R
Molar Specific Heat of an Ideal Gas at Constant Pressure If pressure is constant: U = Q P -W = n C P T - P = n C P T - n R T But U is the same for the same T, whether the volume or pressure is constant. U = Q = Q P W n C T = n C P T n R T C = C P R or C P = C + R = (3/2) R + R C P = (5/2) R
Real Gases: Degrees of Freedom An ideal gas is monatomic, and stores internal energy in translational motion only. Real gases are mostly polyatomic, and store internal energy in translational motion vibrational motion Degrees of Freedom rotational motion Equipartition Theorem: Internal energy is divided equally among the degrees of freedom: ½k B T per degree of freedom per molecule. ½RT per degree of freedom per mole.
Example: Diatomic Molecule A diatomic molecule has 3 translational degrees of freedom. 2 vibrational degrees of freedom. 2 rotational degrees of freedom. 7 total degrees of freedom Internal energy of n moles of diatomic molecules: U = 3n(½RT) + 2n(½RT) + 2n(½RT)= (7/2)nRT translation Molar Specific Heat At Constant olume vibration rotation total C 1 du 1 d 7 7 = = nrt = n dt n dt 2 2 R