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## Ch 21 Temperature, Heat, and Expansion

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**Temperature**A measure of the average kinetic energy of the particles in a substance.**Imagine a pail of warm water and a cup of a hot water.**• A 1 & 2 liter bottle of boiling water.**Temperature is NOT a measure of the total KE of molecules in**the substance.**Temperature Scales**1. Fahrenheit (oF) 2. Celsius (oC) 3. Kelvin (K)**Boiling Point**1. Fahrenheit 212 oF 2. Celsius 100 oC 3. Kelvin 373 K**Freezing Point**• 1. Fahrenheit 32 oF • 2. Celsius 0 oC • 3. Kelvin 273 K**Absolute Zero**Point at which all molecular motion has stopped. We have never reached it, but are very close. Scale is used in engineering.**Rankine Temperature Scale**• Temperature scale having an absolute zero, below which temperatures do not exist, and using a degree of the same size as that used by the Fahrenheit temperature scale. • Absolute Zero corresponds to a temperature of −459.67°F;**Temperature Difference (DT)**Is the primemover or force-like quantity in a thermal system. DT – “Delta T”**Ex: 110 oF inside and 40 oF outside. What is the DT?**• DT = 110 – 40 = 70 Fo**Thermometer**• Instrument used to measure temperature. • Based upon liquid expansion in the tube with respect to temperature.**Converting Temperatures**Fahrenheit to Celsius TC = 5/9(TF – 32o)**Ex: Convert 50 oF to oC**TC = 5/9(TF – 32o) TC = 5/9(50 – 32o) TC = 5/9(18o) TC = 10 oC**Celsius to Fahrenheit**• TF = 9/5(TC)+ 32o**Ex: Convert 20 oC to oF**TF = 9/5(TC)+ 32o TF = 9/5(20)+ 32o TF = 36 + 32o TF = 68 oF**Convert Celsius to Kelvin**Tk = Tc + 273 Tc = Tk - 273**Ex: Convert 72 oF to K**TC = 5/9(TF – 32o) TC = 5/9(72 – 32o) TC = 22.2 oC**Tk = Tc + 273**Tk = 22.2 + 273 Tk = 295.2 K**Heat**• Energy transferred from one body to another due to a DT between them.**Once its absorbed by the 2nd body/material it becomes**internal energy.**Heat is energy in transit.**• Heat flows from high to low temperatures.**Heat will flow out of the body at a higher temperature and**into a body at a cooler temperature.**When the heat flows, the objects are said to be in thermal**contact.**Two things can happen:**• The temperature rises. • The object changes state.**The state in which 2 bodies in physical contact with each**other have identical temperatures. • No heat flows between them**Internal Energy**The energy of a substance due to the random motions of its component particles and equal to the total energy.**Quantity of Heat**• When heat is absorbed it raises the temp. or when it’s lost it lowers the temperature.**calorie**• The amount of heat energy required to raise the temperature of 1 gram of water 1 oC.**1 kilocalorie (1000 calories) is used in rating food.**• Written as Calorie (capital C)**Both are units of energy.**• 1 calorie = 4.187 J • BTU – British Thermal Unit (English Unit)**Fuels are rated by how much heat is given off when a certain**amount is burnt.**Heat Transfer**• Specific Heat (Cp) • amount of energy required to raise the temp. of 1 kg of material by 1 degree Kelvin • units: J/(kg·K)or J/(g·°C)**50 g Al**50 g Cu Heat Transfer • Which sample will take longer to heat to 100°C? • Al - It has a higher specific heat. • Al will also take longer to cool down.**– Q = heat loss**+ Q = heat gain T = Tf - Ti Heat Transfer Q = m T Cp Q: heat (J) m: mass (kg) T: change in temperature (K or °C) Cp: specific heat (J/kg·K or J/g.oC)**Coffee cup Calorimeter**Heat Transfer • Calorimeter • device used to measure changes in thermal energy • in an insulated system, heat gained = heat lost**Specific Heat (c)**• Is the quantity of heat required to raise the temperature of a unit mass of that substance by 1oC.**Units of Specific Heat**• Joules per kilogram-Celsius degree • J/kg-Co**Specific heat of water is 4190 J/kg-Co**• On Pg. 220 is a table of Specific heat for different substances.**Heat Gain or Loss**Q = mcDT**Q = quantity of heat**• m = mass of the substance • c = specific heat of the substance • DT = Temperature Difference**Ex : A 0.5 kg cast iron skillet is heated from 20 Co to 55**Co. How much heat is was absorbed by the iron?**m = .5 kg**c = 448 J/kg-Co DT = (55 – 20) = 35 Co Q = ?**Q = mcDT**Q=(.5 kg)(448 J/kg-Co) (35Co) Q = 7840 J**Ex: A 1 kg of lead at 100 oC is dropped into a bucket**containing 1 kg of water at 0 oC. What is the final temperature of lead and water when it reaches equilibrium?**We know the heat lost by the lead is gained by the water.**Qlead = Qlost = ? m = 1 kg c = 128 J/kg- oC DT = (100 oC – TF)