Topic 11: Energy in Thermal Processes(11.1 - 11.4)

Conceptual #1: Ice Cube What is the most likely temperature of ice-cubes stored in a refrigerator's freezer compartment? (a) -10 °C (b) 0 °C (c) 5 °C (d) It depends on the size of the ice-cubes

•(a) -10 °C -But why?, At 0 °C we can have a water mixture, we want to pull all of the energy out so it is colder, We want no margin, so it is for sure an ice cube, We want a temperature below the melting point, not at it

Conceptual #2 and #3: Ken's Ice-Cubes 2. Ken takes six ice-cubes from the freezer and puts four of them into a glass of water, (He leaves two ice-cubes on the bench top--see next problem), He stirs and stirs until the ice-cubes are much smaller and have stopped melting, What is the most likely temperature of the water at this stage? (a) -10 °C (b) 0 °C (c) 5 °C (d) 10 °C 3. The two ice-cubes Ken left on the bench-top have almost melted and are lying in a puddle of water, What is the most likely temperature of these smaller ice-cubes? (a) -10 °C (b) 0 °C (c) 5 °C (d) 10 °C

2. •(b) 0 °C -But why?, It is in warm water, water is giving thermal energy to it, but it didn't finish melting the ice, It is still in an ice-water mixture phase, so is at 0 °C 3. •(b) 0 °C -But why?, Similar reason as above, the counter is acting in the same way as the water, the ice is still in a phase change, so is still at 0 °C

Heat Example: Breakfast Bar

It is 500 Calories, What is Δh? •Drawing of a person pulling on a pulley system holding a 1000 kg weight, What is the Δh of the weight? -ΔE = 0 -ΔEchem + ΔEg = 0 -ΔEchem + mgΔh = 0 -Δh = -ΔEchem / mg -[(-500 Cal)(1000 cal / 1 Cal)(4.186 J / 1 cal)] / [(1000 kg(9.8 m/s²)] -Δh = 214 m

Latent Heat and Phase Change

Phase Change Diagram from above •Temperature, ΔT(°C) on y-axis vs. Heat Energy, Energy Added, ΔQ(J) on x-axis -Starts at Low T and Q, Starts as solid, Rises linearly, As heat energy is added, there is no more temperature change for a point(either melting or freezing), Now a liquid, Rises linearly, As heat energy is added, there is no more temperature change for a point(either vaporization or condensation), Now a gas, Rises linearly •1.) ΔQ = mcΔT(between phase transition, e.g., solid, liquid, gas) •2.) ΔQ = mL(during phase transition, e.g., melting, freezing, vaporization or boiling, condensation or condensing) -L = Latent heat -Lfusion = Lf, Lvaporization = Lv, These are just numbers or values for a given substance

Problem 2.1: Melting Ice (a) How much thermal energy must be absorbed by ice of mass m = 720 grams at -10 °C to take it to a liquid state of 15 °C? (b) If we supply the ice with a total energy of only 210 kJ(as heat), what then are the final state and temperature of the water?

Start these problems by asking yourself where we are at on the phase diagram graph •For (a) -Before: Ice(i), mi = 720 g, Ti = -10 °C, ci = 2220 J/kg°C -After: Water(w), mw = 720 g, Tw = 15 °C, cw = 4190 J/kg°C -Latent heat of fusion(melting phase transition), Lf = 3.33 X 10^5 J/kg -ΔQtotalabsorbed = ΔQheatingice(T increasing) + ΔQphasetransition(No change in T) + ΔQheatingwater(T increasing) -Think of the graph, We start with ice at T = 10 °C, T increases to 0 °C, then it starts melting(phase transition) and isn't changing T, then becomes completely water and raises to 15 °C -ΔQ = miciΔTi + miwLf + mwcwΔTw, iw is a mixture of ice and water, they coexist during the phase transition -ΔQ = (0.720)(2220)(0 - -10 °C) + (0.720)(3.33 X 10^5) + (0.720)(4190)(15 - 0), ΔT = Tf - Ti as always, Notice the same amount for m, matter is conserved in this system -ΔQ = 15984 + 239760 + 45252 = 300996 J = ~301 kJ -5% to heat the ice, 80% to complete the phase change from ice to water, and 15% of the energy to heat the water = 100% of the energy required for this process -The inertia or resistance to T change for ice is very small, but notice how much energy is required to break the bonds during the phase change •For (b) -If we add 210 kJ = 210,000 J? -It is enough to warm up the solid to 0 °C, but not enough for a complete phase change -ΔQtotal = ΔQice = ΔQphasechange -210,000 J = 15984 J + ΔQphasechange -ΔQphasechange = 194016 J -It is still in its phase change, not enough energy to get all the way through, it is a mixture of ice and water at 0 °C, Remember, T doesn't change during phase transitions -Fraction into phase = ΔQphasechange/ΔQphasechangetotal = 194016 / 239760 = 0.81 = f -What do you see? 81% water, the rest is ice -mass melted =f(mtotal) = 0.81(720 g) = 583 g of water, So 137 g of ice is left

Phase Change Diagram

Temperature, ΔT(°C) on y-axis vs. Heat Energy, Energy Added, ΔQ(J) on x-axis •Starts out low in Temperature and Heat Energy, solid, atoms in a cubic structure, tightly aligned, Increasing in Temperature and Heat Energy •As we keep adding Heat Energy, it gets to a point where it no longer increases in Temperature, If adding Heat Energy, it is Melting(solid to liquid), If losing Heat Energy, it is Freezing(liquid to solid) •It then becomes a liquid, atoms are further apart but still pretty close, Increasing in Temperature and Heat energy •As we keep adding Heat Energy, it gets to a point where it is no longer increasing in Temperature, If adding Heat Energy, it is Vaporization(liquid to gas), If losing Heat Energy, it is Condensation(gas to liquid) •It then becomes gas, atoms are far apart and floating freely

For this section

We are focused on the changes of matter when we are adding or removing thermal energy to or from a system

Specific Heat and Calorimetry

We have two gases, A and B, TA < TB •We reach Thermal Equilibrium, TA = TB -ΔQA increased, ΔQB decreased -ΔQA = +5 J, then ΔQB = -5 J, Net Energy = 0 J -ΔQA + ΔQB = 0 -ΔQA(Energy gained) = -ΔQB(Energy lost) ΔQ is made up of 3 things, C(specific heat capacity), m, and ΔT •ΔQ = mcΔT, "M-CAT" •Example: Heating Alcohol -ΔQ = 100 J, c = 2000 J/kg/°C, m = 200 g = 0.200 kg, What is ΔT? -ΔQ = mcΔT -ΔT = ΔQ / mc -ΔT = (100 J) / (0.2 kg)(2000 J/kg/°C) -ΔT = (1 X 10^2) / (2 X 10^-1)(2 X 10^3) = 1/4 °C ΔT = ΔQ / mc •Temperature, ΔT(°C) on y-axis vs. Mass, m(kg) on x-axis -As temperature decreases, mass increases exponentially -There is a table in the book with the specific heat capacities(c) of different substances

Conceptual #5: Pam and the Freezer Pam asks a group of friends: "If I put 100 grams of ice at 0 °C and 100 grams of water at 0 °C into a freezer, which one will eventually lose the greatest amount of thermal energy? (a) Cat says: "The 100 grams of ice" (b) Ben says: "The 100 grams of water" (c) Nic says: "Neither because they both contain the same amount of thermal energy" (d) Mat says: "There is no answer, because ice doesn't contain thermal energy" (e) Jed says: "There's no answer because you can't get water at 0 °C" Which of her friends do you most agree with?

•(b) Ben says: "The 100 grams of water" -But why?, Let's look at the phase diagram, At 10 °C, ice in a freezer, it is at the incline point, At 0 °C, ice is just starting the phase change, Water at 0 °C has much more thermal energy, it loses way more energy in the transition(same T but further to the right on the graph), mcΔT is relevant during the inclines, mL is during the phase changes, different values for c because of quantum mechanics, which we won't go into why cwater>cice>cwater vapor

Conceptual #7: Weather Forecast A group is listening to the weather forecast on a radio, They hear: "....tonight it will be a chilly 5 °C, colder than the 10 °C it was last night..." (a) Jen says: "That means it will be twice as cold tonight as it was last night" (b) Ali says: "That's not right, 5 °C is not twice as cold as 10 °C" (c) Raj says: "It's partly right, but she should have said that 10 °C is twice as warm as 5 °C" (d) Guy says: "It's partly right, but she should have said that 5 °C is half as cold as 10 °C" Whose statements do you most agree with?

•(b.) Ali says: "That's not right, 5 °C is not twice as cold as 10 °C" -But why?, Celsius is not an absolute scale, The better way to think of it is in Kelvin, it has an absolute value of 0 to compare everything else to, to measure how hot or cold something is, So if it were between 5 K and 10 K, it would be true, but 5 °C and 10 °C is 278 K and 283 K, which is well under a 10% increase

Conceptual #4: Cups of Water Lee takes two cups of water at 40 °C and mixes with one cup of water at 10 °C, What is the most likely temperature of the mixture? (a) 20 °C (b) 25 °C (c) 30 °C (d) 50 °C

•(c) 30 °C -But why? We can do the math for it or reason through, because there is more of the warmer water, the equilibrium temperature would be closer to 40 °C than 10 °C, so is the only plausible option, Otherwise you can do 40 °C + 40 °C + 10 °C / 3 substances = 30 °C total

Conceptual #6: Mel and Boiling Water Mel is boiling water in a saucepan on the stovetop, What do you think is in the bubbles that form in the boiling water? Mostly: (a) air (b) oxygen and hydrogen gas (c) water vapor (d) There's nothing in the bubbles

•(c) Water vapor -But why?, It is in its transition from liquid to gas, this gas forms in the liquid water causing the bubble, then it pops and releases into the air

HW #6 A thermos contains 150 cm³ of coffee at 85 °C, To cool the coffee, you drop two 11-g ice cubes into the thermos, The ice cubes are initially at 0 °C and melt completely, What is the final temperature of the coffee? Treat the coffee as if it were water

•A cup of coffee is drawn, it is giving energy in the form of heat to the ice cubes, There are three objects involved, a thermos(1), the coffee(2), and the two ice cubes(3) -V2 = 150 cm³, T2 = 85 °C, c2 = 4186 J/kg°C -m3 = 11 g = 0.011 kg, T3 = 0 °C, L3 = 33.5 X 10^4 J/kg •ΔQ = 0 -ΔQcofee(negative) + ΔQice(positive) = 0 •ΔQcoffee = m2c2ΔT -Where m2 = ρ2V2 = (1 g/cm³)(150 cm³) = 150 g = 0.150 kg -ΔQcoffee = (0.150)(4186)(Tf - 85) = 627.9(Tf - 85) •ΔQice = 2(m3L3 + m3c3ΔT), there are two ice cubes and two expressions, energy needed to melt and to heat up -ΔQice = 2[(0.11)(33.5 X 10^4) + (0.11)(4186)(Tf - 0)] -ΔQice = 7.37 X 10^3 + 9.21 X 10^1 Tf(positive) = 2 [3685 + 46.05 Tf] = 7370 + 92.09 Tf •ΔQcoffee = -ΔQice -627.9(T - 85) = -7370 - 92.09 T, where Tf is T -A(T - 85) = -B - CT, were A = 627.9, B = 7370, and C = 92.09 -AT - 85A = -B - CT -AT + CT = 85A - B -(A + C)T = 85A - B -T = (85A - B) / (A + C) -T = [85(627.9) - 7370] / [627.9 + 92.09] -T = (4.600 X 10^4) / (719.99) = 64 °C

HW #3 An ice chest at a beach party contains 12 cans of soda at 5.0 °C, Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/kg°C, Someone adds a 6.5-kg watermelon at 27 °C to the chest, The specific heat capacity of watermelons is nearly the same as that of water, Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon

•Ice Chest -Drawing of the watermelon transferring energy as heat to the soda cans -Soda: Ts = 5 °C, ms = 0.35 kg, cs = 3800 J/kg°C -Watermelon, Tw = 27 °C, mw = 6.5 kg, cw = 4186 J/kg°C -The chest does not matter •Find final temperature of the chest •ΔQsystem = 0 = 12 ΔQsoda + ΔQwatermelon = 0 •ΔQsoda = mc(Tf - Ti) -(0.35)(3800)(T - 5) = 1330(T - 5), where Tf is T •ΔQwatermelon = mc(Tf - 27) -(6.5)(4186)(T - 27) = 27209(T - 27) •So... -12 ΔQsoda = -ΔQwatermelon -12(1330) (T - 5) = -27209 (T - 27) -15960 (T - 5) = 27209 (27 - T) •We will call 15960 "A" and we will call 27209 "B" •AT - 5A = 27B - BT -AT + BT = 27B + 5A -(A + B)T = 27B + 5A -T = (5A + 27B)/(A + B) -T = [5(15960) + 27(27209)] / [15960 + 27209] -T = 19 °C

Heat and Internal Energy

•Internal energy(U): Random motions between particles of substances -Connected to temperature -Total energy within a system •Heat/heating(Q): Process in when ΔT increases or decreases -Heat is to Thermal Energy as Work is to Mechanical Energy -1 calorie = 4.186 Joules(J) -1 food Calorie = 1000 calories

HW #5 The latent heat of vaporization of sweat at body temperature(37 °C) is 2.42 X 10^6 J/kg, To cool the body of a 75-kg jogger(average specific heat capacity = 3500 J/kg°C) by 1.5 °C, how many kilograms of water in the form of sweat have to be evaporated?

•Jogger -Drawing of a guy, droplet of sweat on his arm, it releases heat from the body, body cools(1) and sweat comes off(2) -Lv water at 37.5 °C = 2.42 X 10^6 J/kg -Mjogger = 75 kg -Cjogger = 3500 J/kg°C -ΔT(cool) = -1.5 °C -Find Msweat •ΔQ = ΔQ1 + ΔQ2 = 0 -MbodyCbodyΔTbody + Msweat + Lsweat = 0 -Msweat = -MbCbΔTb / Lv -Msweat = -[(75 kg)(3500 J/kg°C)(-1.5 °C)] / [2.42 X 10^6 J/kg] = 0.163 kg = 163 g

HW #1 You are sick and your temperature is 312.0 kelvins, Convert this temperature to the Fahrenheit scale

•Sick! •Tk = 312 K, Find Tf(temperature in Fahrenheit) -Tk - Tc + 273.15 -Tc = Tk - 273.15 = 38.85 °C -Tf = (9/5)Tc + 32, where 9/5 is for the expanded scale and 32 is the ice point -Tf = (9/5)(38.85) + 32 = 101.93 °F, Yep, you're sick!

What is specific heat capacity really?

•Similar to Newton's Second Law -F = ma, F is the cause of motion, a is the effect of motion -a = F/m, m is the resistance to acceleration, or changes in its motion or velocity -If m increases by a lot, then a decreases by a lot, i.e., heavier objects are more difficult to change their motion •ΔQ = mcΔT -ΔQ is the cause of temperature change, ΔT is the effect -ΔT = ΔQ/(mc), mc is the resistance to temperature changes -From ice to water, it get a higher value for specific heat capacity(c), so it is harder to change T •Latent heat has to do with what happens during the phase transitions -Lf and Lv, the energy required(in J/kg) to pass through the transition, We will have to write these values of different substances on our notecards for our quizzes and exams

HW #4 When resting, a person has a metabolic rate of about 3.0 X 10^5 joules per hours, The person is submerged neck-deep into a tub containing 1.2 X 10^3 kg of water at 21.00 °C, If the heat from the person goes only into the water, find the water temperature after half an hour

•Submerged in a tub -Drawing of a person in a tub of water of mass m, their is heat loss from the person to the water -Metabolic rate = R = 3.0 X 10^5 J/hr -Mass of the water = M = 1.2 X 10^3 kg -Tw = 21.00 °C -Find Tw after 1/2 an hour •ΔQ = ΔQ1 + ΔQ2 = 0 -ΔQ = -RΔt + M(cw)ΔT = 0 -ΔT = RΔt / M(cw) -ΔT = [(3.0 X 10^5 J/hr)(0.5 hr)] / [(1.2 X 10^3 kg)(4186 J/kg°C)] = 0.030 °C •Tf = Ti + ΔT = 21.00 °C + 0.030 °C = 21.03 °C, a small increase

Explanation of Phase Change Diagram Parts

•T vs. ΔQ, it can be positive or negative, we are adding or taking away energy from the system -We start with ice, we add heat, it increases in temperature until 0 °C -As we keep adding energy in the form of heat, it stays at 0 °C, No T change(energy is not going into the random motion of the molecules), Latent heat of fusion(Lf), This is because at the microscopic level, this energy is going into breaking the bonds(IMFs, etc.) of ice into water, making it more loosely organized -Once it is all broken down into water, adding heat energy will raise its temperature again until at 100 °C -There is no temperature change, Latent heat of vaporization, energy is going into breaking the water molecules into their gas phase, Once all gas, T will start to increase as ΔQ increases

HW #2 A thin rod consists of two parts joined together, One-third of it is silver and two-thirds is gold, The temperature decreases by 26 °C, Determine the fractional decrease in the rod's length, ΔL / (L0silver + L0gold), where L0silver and L0gold are the initial lengths of the silver and gold rods

•Two-Part Rod -Drawing of a rod, 1/3 is gold, 2/3 is silver •ΔT = -26 °C •Ag(silver) -L1 = 1/3 length units -α1 = 19 X 10^-6 °C^-1 •Au(gold) -L2 = 2/3 length units -α2 = 14 X 10^-6 °C^-1 •Find fractional decrease in length: ΔL / L0 •ΔL = ΔL1 + ΔL2 = α1(L1)(ΔT) + α2(L2)(ΔT) -ΔL = α1(1/3 L)(ΔT) + α2(2/3 L)(ΔT) -ΔL = (1/3 α1 + 2/3 α2)(L)(ΔT) •ΔL/L = (1/3 α1 + 2/3 α2)(ΔT) -ΔL/L = [1/3(19^10^-6) + 2/3 (14 X 10^-6)][-26 °C] -ΔL/L = [(3.17 X 10^-6) + (9.33 X 10^-6)][-26 °C] = 0.000325 = ~0.04%

ΔQ = mcΔT

•Units: (J) = kg(J/kg°C)(°C) -They cancel out and we are left with joules, the unit of energy that is being added or taken •c is specific heat capacity, It is a number, It depends on the phase the substance is in -It may also sometime be written in J/kgK, but as we know, ΔTK = ΔT°C, so it doesn't matter

Problem 2.2: Copper Penny A copper penny whose mass mc is 75 g is heated in a laboratory oven to a temperature T of 312 °C, The penny is then dropped into a glass beaker containing a mass mw = 220 g of water, The initial temperature Ti of the water and the beaker is 12 °C, Find the final temperature Tf of the system at thermal equilibrium Assume: •The specific heat capacity of the glass beaker is 840 J/kg°C •The penny, beaker, and water are an isolated system and the water does not vaporize Not listed: Mass of glass beaker = 224 g, Temperature of glass beaker = 12 °C

•c for copper penny, mc = 75 g, Tc = 312 °C, cc = 386 J/kg°C •w for water, mw = 220 g, Tw = 12 °C, cw = 4190 J/kg°C •b for glass beaker, mb = 224 g, Tb = 12°C, cb = 840 J/kg°C -The glass beaker does matter, there are 3 objects in contact with each other, the coin is transferring thermal energy as heat, All come to the same equilibrium temperature, eventually there are no more energy transfers because they are all the same temperature •Find the final equilibrium temperature(Tf) •ΔE = 0, If system is isolated, The conservation of energy will help us here -E = thermal energy -ΔQ = 0 -ΔQcopper + ΔQwater + ΔQbeaker = 0 -ΔQc + ΔQw + ΔQb = 0 -We expect ΔQc to decrease and ΔQw and ΔQb to increase, with ΔQtotal being 0 •Phase change? -No, No(we are told the water doesn't vaporize), and No, Obviously we don't expect any phase changes of any of the objects, so we don't need to worry about latent heats, The solid coin is getting colder so obviously won't melt, the water won't vaporize, and the glass beaker won't melt •mc(cc)(ΔTc) + mw(cw)(ΔTw) + mb(cb)(ΔTb) = 0 -Tc will decrease, Tw will increase, and Tb will increase, Remember ΔT is just Tfinal - Tinitial •So... -mc(cc)(Tcf - Tci) + mw(cw)(Twf - Twi) + mb(cb)(Tbf - Tbi) = 0 -T = Tf = Tcf = Twf = Tbf = ?, This is the Final Equilibrium Temperature we are looking for -T0 = Twi = Tbi = 12 °C •Now... -mc(cc)(T - Tci) + mw(cw)(T - T0) + mb(cb)(T - T0) = 0 -mc(cc)(T) - mc(cc)(Tci) + mw(cw)(T) - mw(cw)(T0) + mb(cb)(T) - mb(cb)(T0) = 0 -mc(cc)(T) + mw(cw)(T) + mb(cb)(T) = mc(cc)(Tci) + mw(cw)(T0) + mb(cb)(T0) -T[mc(cc) + mw(cw) + mb(cb)] = mc(cc)(Tci) + mw(cw)(T0) + mb(cb)(T0) -T = [mc(cc)(Tci) + mw(cw)(T0) + mb(cb)(T0)] / [mc(cc) + mw(cw) + mb(cb)] -We know everything else except T, so can plug them in -T = [0.075(386)(312) + 0.220(4190)(12) + 0.224(840)(12)] / [0.075(386) + 0.220(4190) + 0.224(840)] -T = (22350) / (1139) = 19.6 °C = ~20 °C •Knowing ΔQ = 0 and Tf = 19.6 °C(equilibrium temperature), and that ΔT = Tf - Ti, we can use this as a check -ΔQc = mc(cc)ΔT = 0.075(386)(19.6 - 312) = -8465 J -ΔQw = mw(cw)ΔT = 0.220(4190)(19.6 - 12) = +7006 J -ΔQb = mb(cb)ΔT = 0.224(840)(19.6 - 12) = 1430 J -ΔQ = (-8465 J)+(7006 J)+(1430 J) = 0 J ✓

Latent Heat and Phase Change Example: Ice Cube

•m = 1 kg, ΔT = 10 °C(0 °C to 10 °C), What is Q? -ΔQtot = ΔQmelting + ΔQliquid -ΔQtot = mLf + mcΔT -ΔQtot = (1 kg)(3.33 X 10^5 J/kg) + (1 kg)(4200 J/g°C)(10 °C) -ΔQtot = 333 kJ + 42 kJ = 375 kJ