Therm-1 Thermal Properties Temperature What is temperature? It is a measure of the amount of "atomic jiggling". When something is hot (has a high temperature), its atoms are jiggling a lot. When it is cold (has a low temperature), its atoms are jiggling little. gas at lo T gas at T = 0K gas at hi T As temperature falls, atoms jiggle less and less. At "absolute zero" T = 0 K, all atoms stop, no motion. Temperature T = measure of energy per atom Various temperature scales: FCKC9TT325T T 273.15=+=+ oF = Fahrenheit oC = Celsius K = Kelvin 1 oC = 1 K , 1 oF = (5/9) Co room temperature = 72oF = 22oC = 295 K ≈ 300 K absolute zero = 0 K = –273oC = – 459oF [In the ideal gas law, p V = N k T = n R T (N = #molecules, n = #moles), must always use T in Kelvin.] Thermal energy U = total energy of all atoms (random motion) Heat Q = amount of thermal energy transferred to a body. [Q] = energy, SI unit of heat = joule popular unit of energy = 1 calorie (cal) = 4.184 J Notice calorie spelled with a small "c".Therm-2 1 cal = energy to raise T of 1 gram of water by 1oC 1 kcal = 1000 cal = 1 Cal = 4184 J = "food Calorie" Notice Calorie spelled with a big "C" [ Some primitive cultures use the BTU ("British Thermal Unit) = energy to raise a pound of water by 1oF. 1 BTU = 1060 J ] Definition: heat capacity of an object = heat added per temperature rise (J/K) Definition: specific heat (or specific heat capacity) of a material = c = amount of heat added per unit mass per degree Celsius rise in temperature. If we have a mass m of some material, and we add an amount of heat ∆Q and that produces a temperature rise of ∆T, then specific heat is defined as.. QcmT∆=∆ [c] = J/(kg ⋅ oC) (SI units) Usually write the equation as ∆Q = m c ∆T This equation says that if I have a mass m of some material with specific heat c, and I want to raise its temperature by ∆T, then I have to add an amount of heat ∆Q = m c ∆T. wateroocal kcal Jc 1 1 4186gC kgC kgC== =⋅⋅o⋅ Example: Heat your mug of coffee (which is mostly water) from room temperature to near boiling: m = 200 g, T = 20oC → 90oC. oocal 4186 JQ m c T (200g) 1 (70 C) 14000 cal 14.0 kcal 58600 JgC kcal⎛⎞⎛⎞⎟⎜⎟⎜∆ = ∆ = ⎟ ==×=⎟⎜⎜⎟⎟⎟⎜⎜⎟⎜⎝⎠⋅⎝⎠ Different materials have different specific heats: material c (cal/g⋅C) water 1.00 ice 0.53 aluminum 0.22 dry air 0.24 iron 0.11 (notice that liquid water has a high specific heat compared to other materials)Therm-3 Example: Suppose we have 2 objects, labeled A and B (water and steel, say), with object A hotter than object B. They initially have temperatures TA and TB . (hot) (cold) TA > TB A B mA , TA , cA mB , TB , cB Bring A and B together, allowing them to exchange heat with each other, but not with the outside world ⇒ A will cool, B will heat and both will reach same final temperature Tf . Object A will lose heat: ∆QA < 0 Object B will gain heat: ∆QB > 0 ABAA A BB BAA f A BB f BfAA BB AAA BBQQmc T mc Tm c (T T ) m c (T T )T(m c mc ) m c T mcT∆ = −∆∆ = −∆− = −−+= +BCan have a mixtur …solve for Tf (does not matter if T is in Celsius or Kelvin, but must be consistent). Phase changes. phase = solid, liquid, or gas (S, L, or G) S ↔ L (freezing/melting) or L ↔ G (boiling/condensing) or S ↔ G (sublimation) Solid water (ice) can have any temperature in the range -273oC < T ≤ 0oC Liquid water can have any temperature in the range 0oC ≤ T < 100oC e of ice and water both at T = 0oC If heat is added to the mixture at T = 0oC , some ice melts, but T stays at 0oC until all the ice has melted. Latent heat or heat of transformation = heat required to cause phase change Latent heat of solid/liquid trans. LSL = heat needed to melt 1 g of ice at 0oC. ice + water LSL (water) = 79.7 cal/g Requires 80 cal to melt a single gram of ice, but only 1 cal to raise temp of the liquid by 1oC.Therm-4 Example: How much heat required to change 100 g of ice at T = –10oC into liquid water at T = +10oC? 1. Heat ice to T = 0oC ∆Q1 = m cice ∆T 2. Melt ice at T = 0oC ∆Q2 = m Lf 3. Heat water to Tfinal ∆Q3 = m cwater ∆T ∆Qtotal = 100(0.5)(10) + 100(80) + 100(1)(10) = 500 cal + 8000 cal + 1000 cal = 9500 cal (heat ice) (melt ice) (heat water) Note that most of the energy went into melting the ice because of the large latent heat of water/ice transformation. This is good for people in Boulder. If Lf was not large, we would have big floods every spring, because all the snow would suddenly melt as soon as the temperature rose above melting. LLG = heat of vaporization = heat needed to transform 1 g of liquid water into vapor at 100oC. LLG (water) = 539 cal This is a very large amount of heat ⇒ very expensive to distill water. Example: Tiger, tiger, burning bright... In the 1956 science fiction movie, Forbidden Planet, Captain Adams (played by Leslie Nielsen) vaporizes a tiger with one shot from his "blaster pistol". This tiger is only about 6 meters away from the captain. About how much energy is required to vaporize a tiger? Is it a good idea to release this much energy this close to you? A tiger is mostly water and has a mass of about m = 250 kg (three time the mass of a man). In order to make the tiger boil away, you have to first raise the temperature of the tiger (water) from T = 30oC (healthy tiger temp.) to 100oC (boiling). Then you have to evaporate the water at T = 100oC. For each gram of tiger, the first step requires 70 cal ( = m c ∆T ), and the second step requires 539 cal ( = m LLG ), so let's say, roughly, at least 600 cal is needed per gram. 81000 g 600cal 4JQ 250 kg 6 10 Jkg g cal=×××=× (just a rough calculation so 1 cal ≈ 4 J.) How much energy is this 6×108 J? This energy is about 200 kW⋅hr. [One kilowatt⋅hour is 1000W (ten 100W light bulbs on) for 1 hour.] The power company charges about $20 for this much energy (at 10 cents per kW⋅hr). This energy is also the energy content …