Heat & Thermo Important Shtuff 2015 D Taylor Heat & Thermodynamics Or, Lets try to remember all that stuff we learned last year, K! Mini-LAB 100 Revs. MEASURE T MEASURE T. -Record on WhiteBd -Discuss Basic LAWS 0 th Law 0 th Law Two systems individually in thermal equilibrium with a 3 rd system are in equilibrium with each other. Sorta like If A=B & B=C, then A=C 13-2: Thermal Expansion 13-4 Gas Laws Ideal Gas Laws Boyles Law (1700) P 1 V 1 =P 2 V 2 Charles Law V 1 /T 1 =V 2 /T 2 Guy-Lusacs Law P 1 /T 1 =P 2 /T 2 Put them all together and Which means? 13-4: Kinetic Theory Basic Stuff Kinetic Theory Look on pg 452+ for pix & examples U 13-4: Derivation of U AP2 Ch14 Heat & Heat Transfer EZ PZ 14.2 Temperature Change and Heat Capacity Heat transfer : Fig 14-4 Calorimetry Since heat, Q, is a form of E Q L = Q G Q Total = 0 = mc T So, Calorimetry Q L = Q G Phase Change So, 14.4: Heat Transfer Methods Ch15: Thermodynamics Or, Lets try to remember all that stuff we learned last year, K! Figure 15.3 WRONG! Basic LAWS 1 st Law 1 st Law U f QW Internal energy of a system changes from an initial value U i to a final value U f due to heat Q and work W: U U f U i Q + W U = U f U i = Q + W Q>0 0 when system gains heat & 0 0 when work is done ON system & 0 Q OUT < 0 W ON > 0 W BY < 0 Using the 1 st LAW 2500J of heat is added to a system, and 1800J of work is done on the system. Find change in internal energy of the system? Using the 1 st LAW Work done on a system in one cycle? T of 3 mol is reduced from 540K to 350K by adding 5500J of heat to it. Find U & W done by or on gas. T of 3 mol is reduced from 540K to 350K by adding 5500J of heat to it. Find U & W done by or on gas. by or on gas? Using the 1 st LAW Work done on or by a gas? Basic LAWS 2 nd Law 2 nd Law Heat flows from a substance at higher temp to a substance at lower temp and does NOT flow in the reverse direction. Ever. Carnot (IDEAL) Engine Basic LAWS 3rd Law 3rd Law Absolute Zero is impossible to reach in a finite number of steps.