Final Exam. Wednesday, December 10. 1:30 4:30 pm. University Centre Rooms

Transcription

1 Final Exam Wednesday, December 10 1:30 4:30 pm University Centre Rooms questions, multiple choice The whole course, equal weighting Formula sheet provided 26

2 Lab and Tutorial Marks Final marks will be put on the web by Friday, December 5 Check your marks! Report errors to Dr. Kunkel (402G Allen) or your TA Thermal Conductivity Read about thermal conductivity in chapter 13 (13.2) for the last lab This is what you need for the final exam 27

3 Specific Heat The amount of heat, Q, to raise the temperature of a mass m of a substance by ΔT o C is: Q = mcδt c = specific heat capacity (or specific heat) in J/(kg.C o ). 28

4 12.39 Blood carries excess energy from the interior to the surface, where energy is dispersed. While exercising, 0.6 kg of blood flows to the surface at 37 o C and releases 2000 J of energy. Find the temperature at which blood leaves the surface. Specific heat of blood = 4186 J/(kg.C o ) The blood loses 2000 J of energy and cools, Q = 2000 J:!T = Q mc = 2000 J = 0.8 C So, blood returns at = 36.2 o C 29

5 Calorimetry Heat is a flow of energy, so should be included in the conservation of energy equation. Energy is conserved, no matter what its form. Calorimetry: studies the flow of heat from one object to another. Calorimeter a thermally insulated container no flow of heat to or from outside. Measure specific heat of an unknown material by heating or cooling to a known temperature, putting into the calorimeter full of liquid of known specific heat, and measuring the equilibrium temperature. 30

6 A calorimeter is made from 0.15 kg of aluminum and contains 0.2 kg of water. Initially, the calorimeter and water are at 18 o C. A 0.04 kg mass of unknown material is heated to 97 o C and added to the water. At equilibrium, everything is at 22 o C. What is the specific heat of the mass? Calculate the heat flow into each object, set the sum to zero: m Al c Al!T Al + m H2O c H2O!T H2O + m c!t = 0 Al water unknown So, c = m Al c Al!T Al m H2O c H2O!T H2O m!t c = (22 18) (22 18) 0.04 (22 97) = 1296 J/(kg.C ) 31

7 12.40 A piece of glass is at 83 o C. An equal mass of liquid at 43 o C is poured over the glass. An equilibrium temperature of 53 o C is reached. Assuming negligible heat loss, find the specific heat of the liquid. Specific heat of glass = 840 J/(kg.C o ) If no loss of heat: mc glass!t glass + mc liquid!t liquid = 0 That is, 840(53 83) + c liquid (53 43) = 0 c liquid = = 2520 J/(kg.C ) 32

8 Latent Heat: Change of Phase The three phases of matter: gas, liquid, solid. Heat is absorbed, or released, when melting/freezing or boiling/ condensation occurs, and temperature remains constant during the change. Latent heat: the energy absorbed or released during a phase change. 33

9 Latent Heat Heat absorbed/released, Q = ml, L = latent heat. Melting/freezing: Latent heat of fusion L f = heat absorbed/released per kilogram on melting/freezing. Boiling/condensing: Latent heat of vaporization L v = heat absorbed/released per kilogram on boiling/condensing. Water: latent heat of fusion = J/kg latent heat of vaporization = J/kg 34

10 12.56 The latent heat of vaporization of water at body temperature is J/kg. To cool the body of a 75 kg jogger (average specific heat = 3500 J/(kg.C o )), by 1.5 o C, how many kilograms of water in the form of sweat have to be evaporated? The vaporization of 1 kg of water requires J of energy. Cooling a mass of 75 kg by 1.5 o C releases an amount of energy equal to: Q = m c!t = = 393,800 J This thermal energy will vaporize a mass m of water: m = 393,800 J J/kg = 0.16 kg of water 35

11 12.58 A 0.2 kg piece of aluminum has a temperature of -155 o C and is added to 1.5 kg of water at 3 o C. At equilibrium, the temperature is 0 o C. Find the mass of ice that has become frozen. Specific heats: Al = 900 J/(kg.C o ) Water = 4186 J/(kg.C o ) Latent heat of fusion of water: J/kg Let m be the mass of water that freezes. Then (1.5 m) kg does not freeze. Heat flows: 0.2 kg of aluminum warms from -155 o C to 0 o C 1.5 kg of water cools from 3 o C to 0 o C mass m of water freezes at 0 o C 36

12 Heat flows: 0.2 kg of aluminum warms from -155 o C to 0 o C 1.5 kg of water cools from 3 o C to 0 o C mass m of water freezes Heat flow: [ ( )] + [ (0 3)] [m ( )] = 0 warm 0.2 kg Al to 0 o C cool 1.5 kg water to 0 o C freeze m kg of water at 0 o C m = kg 37

13 g of liquid water are at 0 o C and another 2 g are at 100 o C. Heat is removed from the water at 0 o C, completely freezing it at 0 o C. This heat is used to vaporize some of the water at 100 o C. How much liquid water remains? L f = J/kg released when water freezes L v = J/kg absorbed when water vaporizes Freeze 2 g water heat 0.002L f is released Heat 0.002L f vaporizes mass m of water Heat released in freezing water = ml f = ( ) = 670 J Mass of water at 100 o C that is vaporized by 670 J of heat is: m = (670 J)/( J/kg) = kg = 0.3 g 1.7 g of liquid water remain. 38

14 12.65 It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinetic energy is converted into heat via friction. Find the minimum speed for this to happen for a bullet at 30 o C. Lead: c = 128 J/(kg.C o ) L f = 23,200 J/kg, melting point o C Heat mass m of lead from 30 o C to o C Melt mass m of lead Heating: Q 1 = m c!t = m 128( ) = 38,054m J Melting Q 2 = m L f = 23,200m J Total heat needed = (38, ,200)m = 61,254m J = 1 2 mv2 v = 350 m/s 39

15 Summary of Temperature and Heat Temperature: T ( o C) = T (K) Thermal expansion: Specific heat: Latent heat:!l = "L 0!T!V = "V 0!T Q = mc!t Q = ml f melting/freezing Q = ml v boiling/condensation Heat flows from high temperature to low 40