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2nd Law of HW 12-2 AND 12-3 Thermodynamics: • Heat always flows from HOT to COLD Heat transfer Khan IR Camera Heat ≠ Temperature Bozeman temp • Temperature : proportional to average kinetic energy of particles (Units: K (Kelvin) ; oC (Celcius) • Heat – energy that flows from a hotter object (higher temperature) to a colder object (lower temperature) (Units: J (joule) WORK HW 12-2 Copy at bottom of p. 128; write equation on top of 129. Q = s×m×ΔT • Q = heat absorbed or lost +Q if absorbed, -Q if lost • m = mass in grams • ΔT = temperature change: Δ T = Tfinal – Tinitial • s = specific heat capacity units of s are J/gºC NO, values range from 5 to 15 50 ÷ 5 = 10 J KINETIC ENERGY #1: 30 ÷ 3 = 10 10 #1: 12 ÷ 3 = 4 4 #1; Higher Average KE HOT TO COLD HEAT WILL FLOW FROM HOT (SYSTEM #1) TO COLD (SYSTEM #2) #1: 21 ÷ 3 = 7 7 #2: 21 ÷ 3 = 7 7 TEMPERATURE ARE EQUAL (SAME AVERAGE KE; THERMAL = HEAT EQUILIBRIUM = BALANCE ; TEMPERATURES ARE EQUAL THERMAL EQUILIBRIUM HW 12-3 CONCEPT OF HEAT Is the amount of material (mass) important in heat transfer? B) The cup will have the higher final temperature; Each individual molecule absorbs more energy $1 $100 When same total amount of energy is distributed amongst a smaller number of molecules, each individual molecule receives more energy. 10 energy units 10 Energy Units split 5 ways split 2 ways Link to video Higher Temp, each molecule has more KE Finish the rest of 12-3 CONTAINER #1 = $500 CONTAINER #1 CONTAINER #2 = $5000 CUP BUCKET OF WATER #1: 24 ÷ 6 = 4 4 4 24 12 #2: 12 ÷ 3 = 4 TEMPERATURES ARE SAME; SAME AVERAGE KINETIC ENERGY CONTAINER #1 HAS STORED MORE TOTAL HEAT 24 vs 12 #1: 300 ÷ 3 = 100 #2: 500 ÷ 10 = 50 100 50 300 500 CONTAINER #1 ; HIGHER AVERAGE KINETIC ENERGY CONTAINER #2 HAS STORED MORE TOTAL HEAT 500 vs 300 Do all objects absorb and store heat in the same way? If both cars are left in the sun in the senior lot for the same amount of time will the interiors be at the same temperature at lunch time? Black Porsche White Porsche Do both of these materials store heat in the same way? Water Rock “s” is Specific Heat Capacity • s = amount of energy needed to raise the temp of 1 g of a substance by 1ºC. • H2O: Rock: s = 4.18 J/gºC s = 0.6 J/gºC • How much energy in J must be added to 1 g of H2O to raise its temp by 1 oC? 4.18 J • How much energy in J must be added to 1 g of rock to raise its temp by 1 oC? 0.6 J HW 12-3, #8 OCEAN (4.18 J/ goC) ROCK (0.6 J/ goC) • MORE energy must be absorbed to raise T. • MORE energy must be lost to lower T. • LESS energy must be absorbed to raise T. • LESS energy must be lost to lower T HW 12-3, #9 COPPER (0.13 J/ goC) ALUMINUM (0.89 J/ goC) • Stores less energy when raising Temp; • Releases less energy when lowering Temp • Stores more energy when raising Temp; • Releases more energy when lowering Temp Calorimeter – an insulated container that does not allow heat to enter or leave ALUMINUM HAS HIGHER HEAT CAPACITY; Al HAS MORE HEAT ENERGY STORED at 100 oC THAN Cu. WATER ABSORBS MORE HEAT FROM Al AND REACHES HIGHER FINAL TEMP. Copy at bottom of p. 128; write equation on top of 129. Q = s×m×ΔT • Q = heat absorbed or lost +Q if absorbed, -Q if lost • m = mass in grams • ΔT = temperature change: Δ T = Tfinal – Tinitial • s = specific heat capacity units of s are J/gºC EXAMPLE PROBLEM • Calculate J of energy required to heat 454 g of water from 5.4oC to 98.6 oC. s of H2O = 4.18 J/g oC. • Q = ? or x • m = 454 g • Ti = 5.4 oC ; Tf = 98.6 oC ; ∆T = Tf - Ti = 98.6 -5.4 =93.2 oC • Q = sm ∆T = (4.18 J/ g oC)(454 g)(93.2 oC) • Q = 1.77 x 105 J