# Calculating Work. Thermodynamics. Isobaric Process. Isochoric (isovolumetric) Work from Graph: Example 1 2/27/2012. Chapter 15

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1 Pressure (X105 N/m2) Pressure (X105 N/m2) 2/27/2012 Thermodynamics Chapter 15 Calculating Work Work = area under Pressure vs. Volume graph W = Fd F = PA W=PAd W = P V Calculus link V 2 W = - p dv V 1 Isochoric (isovolumetric) No change in volume W = 0 Isobaric Process No change in pressure Volume expands and does work WORK = - AREA Pressure vs. Volume of an Ideal Gas Volume (ml) Work from Graph: Example 1 Pressure vs. Volume of an Ideal Gas A gas expands at a constant pressure of 1 X 10 5 N/m 2 (~100 kpa, ~ 1 atm) from a volume of 10 ml to a volume of 35 ml. Calculate the work done by the gas Volume (ml) 1

2 Pressure (N/m2) Pressure (N/m2) 2/27/2012 W = P V W = (1 X 10 5 N/m 2 )(0.035 L L) W = 2500 J Work from Graph: Example 2 A gas is cooled to produce a decrease in pressure. The volume is held constant at 10.0 m 3, and the pressure decreases from 1 X 10 5 N/m 2 to 0.5 X 10 5 N/m 2. Calculate the work done on the gas. Pressure vs. Volume 1.20E E E E E E E Volume (m3) W = P V W = P X 0 W = 0 J (No work is done) Work from Graph: Example 3 A gas expands and increases in pressure as shown in the next graph. Calculate the work done on the gas. (Remember to include the entire area under the graph). Pressure vs. Volume 1.20E E E E E E E Volume (m3) 2

3 Pressure (N/m2) Pressure (N/m2) 2/27/2012 Pressure vs. Volume 1.20E E E E E E E Volume (m3) Example 4 Calculate the work done from 500 to 1000 cm 3 on the following graph. Remember to go all the way down to the x-axis. ANS: Area under the curve Isothermal Systems Isothermal Example A cylinder contains 7.0 grams of nitrogen (N 2 ). W = - p dv = - nrt dv = - nrt dv V V W = -nrt ln V f = -p i V i ln V f = p f V f ln V f V i V i V i a. Calculate the moles of N 2 b. Calculate the work that must be done to compress the gas at a constant temperature of 80 o C until the volume is halved. Pressure vs. Volume of an Ideal Gas 1.20E E E E E E E E E E E E E E-01 Volume (m3) Top Line = at 500K Bottom Line = at 300K 3

4 Pressure (N/m2) 2/27/2012 Work and Cyclic Processes Cyclic process gas returns to its original state Important for studying Steam engine Car engine U = 0 Q = W Work = Area enclosed Pressure vs. Volume 4.50E E E E E E E E E E Volume (m3) Work and Cyclic Processes Heat (Q) Calculate the work shown in the previous graph. 1. Heat Energy transferred from one body to another because of a difference in temperature 2. Cooking a turkey hot oven cooler turkey 3. Extensive Property depends on amount of material (iceberg vs. water) 4. Unit Joules. Calorie calories 1. calorie amount of heat energy needed to raise the temperature of 1 gram of water by 1 degree Celsius (or Kelvin) 2. Not a nutritional Calorie Converting between heat units 1 cal = 4.18 Joules 252 cal = 1 BTU 1054 Joules = 1 BTU 1 nutritional Calorie = 1000 calories (or 1 kilocalorie) 4

5 James Prescott Joule Weight moves paddles Friction from paddles warms water Work falling = Heat paddles Mechanical Equivalent of Heat Heat: Example 1 How high would you have to climb to work off a 500 Calorie ( 500,000 cal) ice cream? Assume you mass 60 kg. (500,000 cal)(4.186j/cal) = 2.09 X 10 6 J Heat = Work Heat = mgh h = Heat/mg h = 2.09 X 10 6 J/(60 kg)(9.8 m/s 2 ) = 3600 m h ~ 11,000 ft Heat: Example 2 A 3.0 gram bullet travels at 400 m/s through a tree. After passing through the tree, the bullet is now only going 200 m/s. How much heat was transferred to the tree? KE = ½ mv 2 KE = ½ mv 2 KE = ½ (0.003 kg)(400 m/s) 2 KE = ½ (0.003 kg)(200 m/s) 2 KE = 240 J KE = 60 J Heat loss = 180 J to the tree The Laws of Thermodynamics 1 st Law Energy is conserved E = W + Q 2 nd Law Natural processes tend to move toward a state of greater disorder Heat goes hot to cold (Clausius) No device converts all heat to work (Kelvin- Planck) S >0 5

6 The Laws of Thermodynamics 3 rd Law The entropy of a pure crystal at absolute zero is zero G = H - T S Important Definitions Thermodynamics Study of the transfer of energy as heat and work System Objects we are studying Isolated System Closed System Open System No mass or energy may leave/enter Only energy may leave/enter, not mass (Earth) Mass and Energy may leave/enter First Law Sign Conventions Heat is added to the system + Heat is lost - Work on the system + Work done by the system - First Law Example J of heat is added to a system. This heat does 1800 J of work on the system. Calculate the change in internal energy. E = W + Q E = 1800 J J E = 4300 J First Law Example J of heat is added to a system. This increase in temperature allows the system to do 1800 J of work. Calculate the change in internal energy. E = W + Q E = J J E = 700 J Adiabatic Systems No heat exchange (Q=0) Fast processes Heat has no time to enter/leave system Car piston Q = 0 For an ideal gas E = W + Q E = 3/2 nrt E = W E = 3nR T 2 6

7 Work: Example 1 In an engine, 0.25 moles of gas in the cylinder expand adiabatically against the piston. The temperature drops from 1150 K to 400K. How much work does the gas do? (the pressure is not constant). Q = 0 E = Q-W E = W E = 3nR T 2 E = 3(0.25 mole)(8.315 J/K-mol)(400 K K) 2 U = 2300 J W = 2300 J Temperature and Internal Energy Temperature measure of the average kinetic energy of individual molecules Temperature Suppose we heat mugs of water from 25 o C to 90 o C. Internal (Thermal) Energy Total energy of all the molecules in an object U = 3 nrt 2 One Mug Two Mugs 25 o C to 90 o C 25 o C to 90 o C Same Temp. Change Same Temp. Change Requires less heat Requires more heat Specific Heat Specific Heat Amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius or Kelvin Unit J/kg o C Symbol = C The higher the specific heat, the more energy needed to raise the temperature Wooden spoon versus a metal spoon Calculating Heat Examples: Al has a specific heat of 0.22 kcal/kg o C Gold has a specific heat of 0.03 kcal/kg o C Which gets hotter sitting in the sun? Water s specific heat: 1.00 kcal/kg o C or 4186 J/kg o C 7

8 Calculating Heat Suppose you immerse a hot pan into a dishpan of water to cool it. Which will experience a greater change in temperature (gain or loss)? Heat lost = -Heat gained Q1 + Q2 + Q3 +. = 0 Q = mc p Calculating Heat T Q = heat (J) m = mass (kg) C p = specific heat (J/kg o C) T = T final T initial Calculating Heat: Example 1 How much heat must be supplied to a gram iron pan (C = 450 J/kg o C) to raise its temperature from 20.0 o C to 100 o C? Q = mc T Q = (0.500 kg)(450/kg o C)(100 o C -20 o C) Q = (0.500 kg)(450/kg o C)( 80 o C) Q = 18,000 J or 18 kj Calculating Heat: Example 2 Suppose the pan is filled with 400 g of water. What would be the total heat? Q = mc T Q = (0.400 kg)(4186/kg o C)(100 o C -20 o C) Q water =134,000 J Q total = 152,000 J or 152 kj Calculating Heat: Example g of tea at 95 o C is poured into a 150-g glass (C = 840 J/kg o C) at 25 o C. What will be the final temperature of the cup/glass? Heat lost = -Heat gained m tea C tea T = -m cup C cup T (0.200 kg)(4186j/kg o C)(T-95 o C) = -(0.150 kg)(840j/kg o C)(T-25 o C) (837)(T-95 o C) = -(126)(T-25 o C) 837T 79,500 = T 963 T = 82,700 T = 86 o C Q tea = - Q cup m tea C tea T = -m cup C cup T 8

9 Calculating Heat: Example 4 A 100-g piece of aluminum (C = 900 J/kg o C) is heated to 100 o C and immersed in 250-g of water at 20 o C. What is the final temperature of the system? Heat lost = -Heat gained m Al C Al T = -m water C water T (0.100 kg)(900j/kg o C)(T-100 o C) = -(0.250 kg)(4186j/kg o C)(T-20 o C) (90)(T-100) = -(1047)(T-25) 90T 9,000 = 20, T 1137T = 29,930 T = 26 o C Q tea = - Q cup m Al C Al T = -m water C water T Phase Changes Definition - A change of state in which energy is absorbed or released without a temperature change. Example: Freezing water 1. Water is cooled from 25 o C to 0 o C (temperature change, not a phase change) 2. Water freezes at 0 o C (no temperature change occurs during freezing, phase change) Phase Change Suppose we are heating ice: All of the heat goes into melting the ice rather than increasing the temperature Temperature will only rise once it is all melted Solid molecules absorb the heat to get moving rather than increasing temperature Melted molecules now move freely Phase Changes: Think about it. 1. Can you get water above 100 o C in a pot if you are cooking? 2. Can you get steam above 100 o C in a pressure cooker or furnace? 3. Can you get ice below 0 o C in a freezer? Heating Curves Boili ng Meltin g 9

10 Latent Heat Heating Curves No Phase Change Q = mc p T Phase Change Q = ml f or Q = ml v Temperature ( o C) Use q = ml v Boili ng Use q = ml f L f = Latent heat of fusion (freezing/melting) L v = Latent heat of vaporization (boiling/condensing) Meltin g Use q = mc p T Heat (Joules) Important Values Here are some important values that we will need: Substance C Steam 2010 J/kg o C Water 4186 J/kg o C Ice 2100 J/kg o C Latent Heat of fusion (water) L f = 3.33 X 10 5 J/kg Latent Heat of vaporization(water) L v = 22.6 X 10 5 J/kg Heating Curves: Example 1 How much energy is required to cool grams of water at 20.0 o C to make ice at 10.0 o C? There are three steps involved here: 1. Cooling the water 2. Freezing the water 3. Cooling the ice down to 10.0 o C Let s look at the graph: Temperature ( o C) Freezing (q=ml f ) Ice cools (q=mc p T) Heat (Joules) Water cools (q=mc p T) Let s look at it step by step 1. Cooling the water Q = mc p T = (0.100 kg)(4186j/kg o C)(0 o C- 20 o C) Q = 837 J 2. Freezing the water Q = ml f = (0.100 g)(3.33 X 10 5 J/kg)= 3.33 X 10 4 J 10

11 Heating Curves: Example 2 3. Cooling the ice down to 10.0 o C q = mc p T = (0.100 kg)(2100j/kg o C)(-10 o C-0 o C) q = 2100 J How much energy does a refrigerator have to remove from 1.5 kg of water at 20 o C to make ice at -12 o C? Lastly, we just add the three heat values together: 837J + 33,000J J = 36,200 J ANS: 660 kj Calorimeter Foam cup Insulated metal container (thermos) Must consider heat absorbed by the calorimeter in the equations Calorimeter: Example 1 A kg sample of a new alloy is heated to 540 o C. It is placed into 400-g of water at 10.0 o C which is contained in a 200-g Aluminum calorimeter. The final temperature of the mixture is 30.5 o C. Calculate the specific heat of the new alloy. Heat lost = - Heat gained m alloy C alloy T = -[m water C water T + m cal C cal T] m water C water T = (0.400kg)(4186 J/kg o C)(30.5 o C 10.0 o C) m water C water T = 34,300 J m cal C cal T = (0.200kg)(900 J/kg o C)(30.5 o C 10.0 o C) m cal C cal T = 3700 J m alloy C alloy T = -[m water C water T + m cal C cal T] m alloy C alloy T = -[34,300 J J] m alloy C alloy T = -38,000 J (0.150-kg)(C alloy) (10 o C 540 o C) = -38,000 J (-79.5 kg o C)(C alloy ) = -38,000 J C alloy = 478 J/ kg o C 11

12 Example g of ice at -20 o C is added to 500 g of soad at 20 o C. a. Calculate the heat required to raise the temperature of the ice to the melting point. b. Calculate the heat required to melt the ice. c. Is there enough heat energy available if the soda cools to 0 o C? d. Calculate the final temperature of the soda We will skip specific heats of gases 1. Conduction 2. Convection 3. Radiation Heat Flow Conduction Conduction The transfer of energy from molecule to molecule through collisions. 1. Warmer (faster) molecules to slower(colder) molecules 2. Thermometers work through conduction. 3. Depends on the area of contact. Conduction: thermometer Conduction Slower (colder) mercury atoms get bumped and accelerat ed by the collisions with the air The warmer, faster moving air molecules collide with the glass and give some of their kinetic energy of motion to the glass and mercury. 4. Air is a poor conductor - molecules are so far apart, far fewer collisions than in a solid. 5. Insulators usually have a number of air pockets (like in home insulation) to slow the heat flow. 6. Cooking on top of a stove is usually conduction since the pan/pot is in contact with a hotter burner element or a flame. 12

13 Conduction Conduction: Example 1 Q = ka(t 1 T 2 ) t L Q t k A T 1 T 2 L = Rate of heat flow (heat/time) = Thermal Conductivity = Area = Change in Temperature = Length Calculate the rate of heat flow through a window 3 m 2 and 3.2 mm thick. The outer and inner temperatures are 14.0 o C and 15.0 o C Conduction: R-values R values commercial measure of the quality of insulation Higher R = more insulation R = length k R Glass 1 Brick 1 Insulation Convection Convection the transfer of hear by the movement of a large volume of air or liquid 1. Hot-air home heating 2. Radiator wars air that it comes in contact with through conduction. Convection takes over as it moves through the room 3. Convection ovens Convection Then convection causes the warm air to expand into the room (the hot air rises) The cold air sinks and comes in contact with the radiator again. A hot radiator warms air through conductio n (contact between the atoms/ molecules 13

14 Radiation Radiation The transfer of heat through electromagnetic waves 1. Light (microwaves, infrared, visible, etc..) 2. The waves give kinetic energy to molecules and cause them to move faster (increase in temperature). 3. Sun 4. Red heat lamps at fast-food places. Radiation Light and other radiation from the sun strike the earth impart some of their energy to the molecules on the earth Earth What type of heat transfer is occurring in each picture? 14

15 Radiation Stefan-Boltzmann Equation Q = e A(T 1 4 T 24 ) t = 5.67 X 10-8 W/m 2 K 4 (Stefan-Boltzmann constant) A = area T = Temperature (kelvin) e = Emissivity Radiation: Emissivity e = Emissivity Value between 0 and 1 1 Black substances Absorb and emit radiation well 0 Shiny substances Do not absorb or emit well Radiation: Example 1 An athlete is sitting in a locker room at is 15 o C. The athlete has a skin temperature of 34 o C and an e value of 0.7. If his surface area is 1.5 m 2, calculate the rate of heat loss. Q = e A(T 1 4 T 24 ) t Q = (0.7)(5.67 X 10-8 W/m 2 K 4 )(1.5m 2 )( ) t Q/ t = 120 W Radiation: Example 2 A ceramic teapot (e=0.70) and a shiny teapot (e = 0.10) filled with tea at 95 o C. Calculate the rate of heat loss of each if the room is 20 o C. Assume each teapot is about 0.05 m 2. Radiation and Sun Angle Get less radiation as the sun becomes less vertical Would you absorb more sun at midday or 6 PM? Would you absorb more sun in Mexico or Ohio? ANS: 22 W, 3.1 W 15

16 Solar constant = 1350 W/m 2 (about 1000 when adjusted for the air) Q = (1000 W/m 2 )ea cos t Sun: Example 1 What is the energy absorption of a person (0.80 m 2, e = 0.70) getting a tan if the sun makes an angle of 30 o with the vertical. Q = (1000 W/m 2 )ea cos t Q = (1000 W/m 2 )(0.7)(0.80m 2 ) cos 30 o t Q/ t = 490 W (wearing light clothes will reduce e) Thermal Pollution Thermally polluting power plants Coal plants Oil plants Nuclear plants Non-thermally polluting power plants Hydroelectric Tidal energy Wind Solar Work: Example 4 Determine the change in internal energy of 1.0 liter of ware at 100 o C when it is fully boiled. It results in a 1671 L of steam at 100 o C. Assume the process is done at atmospheric pressure. U = Q W Q = ml =(1.0 kg)(22.6 X 10 5 J/kg) = 22.6 X 10 5 J W = P V (1 L = 1 X 10-3 m 3 ) W = (1 X 10 5 N/m 2 )(1671 X 10-3 m 3-1X 10-3 m 3 ) W = 1.7 X 10 5 J U = Q W U = 22.6 X 10 5 J 1.7 X 10 5 J U = 21 X 10 5 J (or 2.1 X 10 6 J) Work: Example 2 Determine the change in internal energy of 1.0 liter of ware at 100 o C when it is fully boiled. It results in a 1671 L of steam at 100 o C. Assume the process is done at atmospheric pressure. U = Q W Q = ml =(1.0 kg)(22.6 X 10 5 J/kg) = 22.6 X 10 5 J 16

17 W = P V (1 L = 1 X 10-3 m 3 ) W = (1 X 10 5 N/m 2 )(1671 X 10-3 m 3-1X 10-3 m 3 ) W = 1.7 X 10 5 J U = Q W U = 22.6 X 10 5 J 1.7 X 10 5 J U = 21 X 10 5 J (or 2.1 X 10 6 J) 17

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