Q & the 1st Law. Heat Capacity. Keep em straight! c = Specific Heat Capacity = C m. C = Heat Capacity. Q = ΔE int Q = CΔT.

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1 Q & the 1st Law Heat Capacity Q = ΔE int Q = CΔT Q = mcδt Q = ml f Q = ml v Keep em straight! C = Heat Capacity c = Specific Heat Capacity = C m c' = Molar c water Heat = 4186 Capacity J = C kg n

2 An example of Q=mc ΔT You take 100g of ice at -25 o C and 15g of steam at 120 o C and put them both in an adiabatically isolated container. What will be the equilibrium temperature and state of the system? An example of Q=mc ΔT #2 You take 120g of ice at -25 o C and 15g of steam at 120 o C and put them both in an adiabatically isolated container. What will be the equilibrium temperature and state of the system? Joule s Experiment ΔE int = Q in + W on Joule increased 1lb of water by 1 o F by dropping 772 lbs of weight 1 foot ΔE int = Q in W by 1kg of water changed by 1 o C requires 4186 J de Mechanical int = dq equivalence in + dw of heat on

3 Joule s Experiment de int = dq in + dw on or ΔE int = Q in + W on This is the 1st law of Thermodynamics Work Isowho? Isobaric Isometric Isothermal

4 Example When Dr. Joule releases the gas into the second chamber, what will be the change in temperature? Isowho? Example n=1 mole, P 1 =3.00 atm, V 1 =1.00L, E int1 =456J, P 2 =2.00atm, V 2 =3.00L, E int2 =912J. The gas is heated and is allowed to expand such that it follows a straight-line path on a PV diagram. Find the heat absorbed by the gas, and draw the PV diagram. Heat Capacities of Gases C p vs. C v For solids and liquids C P C V For ideal monatomic gasses: C V = 3 2 nr C P = 5 2 nr

5 Heat Capacities of Gases For ideal diatomic gasses: C V = 5 2 nr C P = 7 2 nr Side Note 99.03% of our air is diatomic Next Note: E int = C V T

6 Example The heat capacity at constant pressure of a certain amount of a diatomic gas is 14.4 J/K. find: n, E int at T=300K, c v, and C v 2nd Law According to Kelvin No system can absorb heat from a single reservoir and convert it entirely into work without additional net changes in the system or its surroundings 2nd Law According to Clausius A process whose only net result is to absorb heat from a cold reservoir and release the same amount of heat to a hot reservoir is impossible

7 Cyclic Process The intake Cyclic Process The compression Cyclic Process The ignition

8 Cyclic Process The power stroke Cyclic Process The exhaust opens Cyclic Process The exhaust stroke

9 The Otto Cycle 1.W on, Q=0, Isobaric 2.W on, Q=0, Adiabatic 3.W=0, Q h, Isometric 4.W by, Q=0, Adiabatic W=0, Q c, Isometric 6.W on, Q=0, Isobaric Work done by the engine Total work done by the engine during the entire cycle: ( ) 0 W 4 W 2 + 2W 1 4 W 4 W Heat Engines ε Get out Put in W by = Q h Q c ε = W by Q h = 1 Q c Q h

10 Simplified Version 2nd Law According to Kelvin, No system can absorb heat from a single reservoir and convert it entirely into work without additional net changes in the system or its surroundings Note on Adiabatic Curves P 1 V 1 γ = P 2 V 2 γ γ = C P C V W Adiabatic = P 2V 2 P 1 V 1 γ 1

11 Note on IsoThermic W 1 2 = P dv = W 1 2 = nrt V 1 Work done nrt V dv V by the gas!!! 2 1 V dv W 1 2 = nrt ln V 2 V 1 Example n=1.00 mol, V 1=25.0 L. It is an ideal monatomic gas. Find T at each point, Q between each point, and the efficiency of the cycle Refrigerators How do they work?

12 Refrigerators 1. The compressor compresses the freon gas to a high pressure such that T Boiling Point > T Current Refrigerators 2. The freon undergoes the exothermic process of condensation as it moves along the condenser coils Refrigerators 3. As the freon condenses it releases heat into the room

13 Refrigerators 4. The expansion valve releases the pressure (retaining the liquid freon) such that T Boiling Point < T Current Refrigerators 5. The liquid freon goes through the endothermic process known as evaporation Refrigerators

14 Fridge efficiency COP = Q C W on What is maximum efficiency? An engine that is reversible Because the entropy change in the universe is zero for a reversible process What does reversible mean? 1. Friction does not add heat to the system 2. Temperature differences within the system are small 3. The process is quasistatic

15 Carnot Cycle ε C = 1 T c T h Example A Carnot engine works between two heat reservoirs at temperatures 300 K and 77 K. Find the efficiency of the engine If Q h is 100J, what is W by? How much heat does it exhaust If you reversed the cycle, what would the COP be? Entropy ds dq in T

16 Entropy ds dq in T For ideal gas ΔS = C V ln T f + nrln V f T o V o Entropy ds dq T ΔS = C V ln T f + nrln V f T o V o For ideal gas During an isothermal process ΔS = nrln V f V o Entropy ds dq T ΔS = C V ln T f + nrln V f T o ΔS = nrln V f V o V o For Ideal gas During an adiabatic process ΔS = nrln V f V o

17 Entropy ds dq T ΔS = C V ln T f + nrln V f T o ΔS = nrln V f V o V o For Ideal gas During an isobaric process ΔS = C P ln T f T o Entropy Entropy All ideal gas ΔS = C V ln T f + nrln V f T o V o ds dq T ΔS = nrln V For isothermal and f adiabatic free V o expansion 2 nd Law: The entropy of the universe never ΔS = C P ln T f decreases For isobaric T o Example moles of an ideal gas (occupying a volume of 40 L at a temperature of 400 K) is allowed to expand during an adiabatic free expansion to twice its initial volume. What is the change in entropy of the gas? What is the change in entropy of the universe?

18 Example A 200 Kg block of ice at 0 o C is placed in a large lake. the temperature of the lake is just slightly higher than 0 o C, and the ice melts very slowly. What is the What is the What is the ΔS of the ice? ΔS of the lake? ΔS of the universe? Example While working in Antarctica, you place a 1 kg cup of boiling hot coffee on an ice burg at 0 o C. What was the change in entropy of the coffee cup? What was the change in entropy of the ice burg? What was the change in entropy of the universe?