Thermal Physics Tutorial

Jack Thermal Physics Tutorial Convert between various temperature scales Converting from basic temperature scales is pretty easy. The three most common scales are Fahrenheit (oF), Celsius(oC), and Kelvin(oK). To convert from Fahrenheit to Celsius, you use this simple formula: Tf = (9/5 * Tc) + 32. Or to go from Celsius to Fahrenheit, you use: Tc = 5/9(Tf - 32). To go from Celsius to Kelvin, you just add 273, or subtract 273 to go from Kelvin to Celsius. To get from Fahrenheit to Kelvin, you have to convert from Fahrenheit to Celsius first, then add 273. Now try some examples: 1. Convert from 56 oF to Celsius. 2. Convert from 305 oK to Celsius. 3. Convert 289 oK to Fahrenheit. 1. Use this equation: Tc = 5/9(Tf - 32) Plug in what you know Tc = 5/9(56- 32) Solve 5/9 (24) Answer: 13.33 oC 3. First use Tc = Tk - 273 to get to Celsius Tc = 289-273 Tc = 16 oC practice problems 1. Convert 27oF to Celsius 2. Convert 77oC to Fahrenheit 3. Convert 37oC to Kelvin 2. Use the equation: Tc = Tk - 273 = 305 - 273 Answer = 32 oC Then use Tf = (9/5 * Tc) + 32 Tf = (9/5 * 16) +32 Answer = 60.8 oF answers 1. 2.78oC 2. 170.6oF 3. 310oK Convert between Joules and Cal and calories Jack Converting from Joules Cal and calories is quite simple. First note that Cal is also written as c, and calories is also written as C. 1 c equals 1000 C. Also, 4.186 joules equals 1 Cal or c. So going from Cal (or c) to joules, you just multiply by 4.186. Or if you need to go from Joules to calories (or C), just divide by 4.186 and then multiply by 1000. Now try some problems to see how well you understand this simple concept. 1. Convert from 34 c to C. 2. Convert from 67 Joules to C. Identify the basic changes of state and their relation to molecular motion Objects can change to 3 different states: solid, liquid, and gas. At solid, objects show very little movement with a low kinetic energy. If you heat a solid, it will then turn into a liquid. A liquid develops medium kinetic energy and shows more movement. Then if you heat a liquid, it turns into a gas. Molecules in this state have a great amount of kinetic energy and move a lot leaving it with no structure. Here is a sketch if you a still a tick confused. Gas Liquid Solid Use phase-changed diagrams This concept is quite simple. Diagramming this is easy. Just remembering what the graph looks like is key. Here is an example problem: Graph a 4 oC solid that changes into a liquid at 122 oC. Here is what you should have: Jack 122oC -4oC Now try some on your own. 1. Graph a solid at 56 oC and changes to 34 oC. 2. Graph ice at -4 oC that heats to a liquid at 23 oC Here is what you should of had: 1. 2. 23oC 56 C 34oC o -4 C o Calculate energy related to specific and latent heats Specific heat is the amount of heat required to raise a 1 kilogram object by 1o Celsius. Each object or substance has a different specific heat. To find the amount of energy transferred to a substance of mass changing its temperature, you use the formula: Q = mcT. Q is the amount of energy transferred in Joules, m is the mass in kilograms, c is the specific heat of the substance and T is the change in temperature in degrees Celsius. Examples What is the amount of energy required to raise the temperature of .5 kg of water by 3oC? Q = mcT Q = (.5) (4186) (3) Q = 6279 J When 6050 Joules of energy are added to 7 kg of copper, what is the change in temperature of the copper? Specific heat of copper = 387. Jack Q = mcT 6050 = (7) (387) T 6050 = 2709T 2709 2.23oC = T Practice problems 1. What is the amount of energy required to raise 34 kg of steam by 10oC? Specific heat of steam is 2010 J. 2. 654321 Joules of energy are added to water, which raises 21oC, what is the mass? Specific heat of water is 4186 J. 3. 78 kg of Lead is at 17oC. If the final temperature is 55oC, how much energy was added? Specific heat of lead is 128 J. Answers 1. 683400 J 2. 7.44 kg 3. 379392 J what about latent heat? Latent heat is basically the same as specific heat. The only difference is that latent heat has two types: fusion and evaporation. Latent heat of fusion is used when a phase change occurs during melting or freezing. Latent heat of evaporation is used when a phase change occurs during boiling or condensing. The equation you use for this is simple: Q = +- mL. Q is the energy. M is mass and L is latent heat. The plus sign is chosen when energy is absorbed by a substance, as when ice is melting. The negative sign is chosen when energy is removed from a substance, as when steam condenses to water. Like specific heat, objects and substances each have their own latent heat of fusion and evaporation. Examples How much energy is required to vaporize 2 kg of liquid helium? Lf = 2.09 x 104 Q = +- mL Q = (2) (2.09 x 104) Jack Q = 41800 J How much energy is required when 10 kg of steam changes to water? Lw = 3.33 x 105 Q = +- mL Q = -10 (3.335) Q = -3330000 J Practice 1. How much energy is required for 15 kg of lead to vaporize? Ll = 8.7 x 105 2. How much energy is required when 30 kg of ice changes to water? Lw = 2.26 x 105 3. How much energy is required when 40 kg of water changes to ice? Lw = 3.33 x 105 Answers 1. 1.3125 x 107 J 2. 6.78 x 106 J 3. 1.332 x 107 J Use Qlost = Qgained to solve for unknown mass or heat When combing two materials, (ex: water and copper), the amount of energy lost by one is equal to the amount of energy gained by the other. So for our example –Qwater = Qcopper. So when we plug what we do know into the equation –m1C1(TF-T1,) = m2C2 (TF-T1,) we can solve for the mass of the object or heat. In the equation m1= mass of first object, C1=specific heat of object 1, Tf= Final temperature, T1= initial temperature, m2= mass of object 2, C2= specific heat of object 2. Example: 300 grams of iron mixes with 450 grams of water at 10 degrees C. Find the initial temperature of the iron if the final temperature is 44 degrees C. m1C1(TF-T1,) = m2C2 (TF-T1,).3 (446)(44- T1) = .45(4186)(44-10) 5887.2-133.8T1 = 64045.8 133.8T1 = 58158.6 T1 = 434.67 degrees C now try some on your own. 1. 545 grams of copper mixes with 100 grams of water at 25 degrees C. Find the initial temperature of the copper if the final temperature is 53 degrees C. Jack Ccopper =387 2. 200 grams of aluminum mix with 300 grams of copper at 78 degrees C. Find the initial temperature of the aluminum if the final temperature is 72 degrees C. Ccopper = 387 Caluminum = 900 answers: 1. 11.57 degrees C 2. 68.13 degrees C Qlost = Qgained to solve for unknown final temperature When combing two materials, (ex: water and copper), the amount of energy lost by one is equal to the amount of energy gained by the other. So for our example –Qwater = Qcopper. So when we plug what we do know into the equation –m1C1(TF-T1,) = m2C2 (TF-T1,) we can solve for the final temperature. In the equation m1= mass of first object, C1=specific heat of object 1, Tf= Final temperature, T1= initial temperature, m2= mass of object 2, C2= specific heat of object 2. Example: 2.34 kg of aluminum at 112 degrees C mixes with 9.98 kg of water at 2.0 degrees C find the final temperature. m1C1(TF-T1,) = m2C2 (TF-T1,) 2.34(900)(112-Tf) = 9.98(4186)(Tf-2) 235872 – 2106Tf = 41776.28Tf – 83552.56 319424.56 = 43882.28Tf Tf= 7.28 35 grams of copper at 78 degrees C mixes with 120 grams of water at 12 degrees C. Find the final temperature. m1C1(TF-T1,) = m2C2 (TF-T1,) .035(387)(78-Tf) = .12(4186)(Tf-12) 1056.51-13.545Tf = 502.32Tf – 6027.84 Jack 515.865Tf = 7084.35 Tf= 13.733 Practice: 120 grams of water at 12 degrees C mixes with 2 kg of water at 99 degrees C. find the final temperature. Answer: Tf= 94.075