Temperature & Heat. Overview

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1 Temperature & Heat Overview

2 Temperature vs Heat What is temperature (degrees)? A measure of the average kinetic energy of the particles in an object What is heat (joules)? That energy transferred between objects because of a difference in their temperatures How does heat flow? Heat flows from a hotter to a colder body by transfer of kinetic energy to adjacent molecules Temperature & Heat 2

3 Units of Measurement There are 3 units of temperature measurement: Degrees Celsius (C) (Water: 0 100) Degrees Fahrenheit (F) (Water: ) Kelvin (K), referencing absolute zero Conversion - Celsius & Fahrenheit F 32 = 9/5*C K = C Temperature & Heat 3

4 Thermometers Analog Digital Measuring Devices Thermocouples The temperature of the junction of 2 different wires creates a small voltage, related to the temperature of the junction Temperature & Heat 4

5 Expansion Each material expands, according to its natural property, as its temperature increases by ΔT degrees C. Options Linear: ΔL = L o α ΔT, where α = coefficient of linear expansion Area: ΔA = A o (2α) ΔT Volume ΔV = V o β ΔT, where Β = coefficient of volumetric expansion Temperature & Heat 5

6 Example A metal bar is 1.6 m long at 21 C. If the bar is heated to 84 C by how much does it expand? ( = 1.7 x 10-5 / C) L = L o * *(T 2 T 1 ) L = 1.6 * 1.7 x 10-5 * (84 21) L = 1.7 x 10-3 m Temperature & Heat 6

7 Specific Heat basic property The amount of heat, in joules, required to raise the temperature of 1 kg of a substance by 1 degree Celsius. Symbol of the specific heat of a substance is c Examples of c Water = 4187 J/kg- C, ie 4187 joules of heat energy are required to raise the temperature of 1 kg of water by 1ºC. Iron = 448 J/kg- C Aluminum = 899 J/kg- C Ice = 2090 J/kg- C Temperature & Heat 7

8 Implications of Specific Heat The heat energy (ΔQ) transferred into or out of a mass (m) of material with a specific heat (c) is given by: ΔQ = m*c*δt, where ΔT is the change in temperature This phenomenon can be used when bodies of different temperatures are brought into contact Hot bodies lose, while cold bodies gain heat Temperature & Heat 8

9 Example of Heat Transfer How much heat is required to raise the temperature of a 2 kg block of copper from 20C to 250C (c CU =386 J.kg/ C)? Solve Q = m*c*δt Q = 2 * 386 * (250 20) Q = 177,560 J Temperature & Heat 9

10 Example What would be the final temperature of a 5kg iron (c = 448) rod at 20C if it received 110,000 J of heat? Solve Q = m*c*δt 110,000 = 5 * 448 * ΔT = ΔT But ΔT = Tf To = Tf 20 Thus, Tf = ΔT + 20 = C Temperature & Heat 10

11 Mixture Examples What happens when 2 bodies of different temperatures are brought into contact, such as a hot piece of metal being placed into a bucket of cold water? The hot body (metal) cools while the cold body (water) warms according to the conservation of energy Heat lost by hot body = Heat gained by cold body, or (m*c*δt) lost = (m*c*δt) gained Heat flows from hot to cold! Temperature & Heat 11

12 See OHP Example Mixtures Temperature & Heat 12

13 Phase Change Change of State All solids, liquids and gases are made of particles, and the only difference between them is how much energy the particles have. If you give the particles energy, or take energy away from them, then you can change their state. Temperature & Heat 13

14 Energy in Energy out Temperature & Heat 14

15 Phase Change Change of State There are 2 different types of heat Sensible (or specific) the addition of which causes a rise in temperature of the material, and Latent the addition of which causes no temperature change until the current state of the material has changed completely, from solid to liquid, liquid to gas or in reverse. Temperature & Heat 15

16 Latent Heat There are 2 different latent heats Solid > liquid = latent heat of fusion or melting ( ex. turning ice to water) Q F = mass * heat of fusion, or Q F = m * h f (where h f = 3.35 x 10 5 J/kg for ice) Liquid > gas = latent heat of vaporization (ex. turning water to vapor) Q V = mass * heat of fusion, or Q V = m * h v (where h v = 2.26 x 10 6 J/kg for water) In graphic form it looks like Temperature & Heat 16

17 Heat can be applied to an object to change its state from solid, through liquid to gas, according to the 5 steps shown: 1. Heat the solid until it reaches melting point (specific heat of solid) 2. Heat the solid until it changes completely to liquid (latent heat fusion) 3. Heat the liquid until it reaches the boiling point (specific heat of liquid) 4. Heat the liquid until it changes completely to gas (latent heat of vaporization) 5. Heat the gas (specific heat of gas) Change of State For example, H 2 O Temperature & Heat 17

18 Heat Phases Water Temperature & Heat 18

19 Steps in Solving Specific/Latent Heat Problems Identify the temperature range in the problem Mark the temperature range on the heat graph On the graph move from the lowest to the highest temperature identifying the sloped/flat stages Calculate the heat (Q) for each step as follows: Sloped > specific heat stage > Q = mcδt Flat > latent heat stage > Q = mh (h for fusion/vaporiz n) Add the Q values for each step to determine the total heat (Q total ) Temperature & Heat 19

20 Example How much heat, in joules, is required to raise the temp of 2 kg of ice from -10 C to 20 C? Mark start and finish temp Count slopes and flats 2 x slopes + 1 flat, meaning: 2 x specific heat segments 1 x latent heat segment Calculate Q for each segment, then total the Q values S F S Temperature & Heat 20

21 Calculations Segment 1: sloped = specific heat for ice being heated from -10 to 0 C Q = mcδt = 2 * 2090 * (0 (-10)) = 41,800 J Segment 2: flat = latent heat converting 0 C to 0 C Q = mh f = 2 * 3.35 x 10 5 = 670,000 J Segment 3: sloped = specific heat for water from 0 C to 20 C Q = mcδt = 2 x 4187 * (20 0) = 167,480 J Total heat (Q) = 879,280 J Temperature & Heat 21

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