Q = CΔT = C(T f T i )

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2 Linear scale : need 2 points to define Fahrenheit [ F] Celsius [ C] Kelvin [K] Conversion factors K C C F Expansion in 1-D T F = 9 T 5 C + 32 T C = T K ΔL = αl 0 ΔT ( ) L = L 0 1+ αδt 0 th Law of thermodynamics -- when the system is in thermal equilibrium, the temperature is constant No Heat flow 1 cal = Btu = J Change of system energy with change of temperature Expansion in 2-D ΔA A 0 ( 2α )ΔT A 0 β A ΔT Q = CΔT = C(T f T i ) Expansion in 3-D β A = 2α β V = 3α ΔV = V 0 ( 3α )ΔT = V 0 β v ΔT C = heat capacity

3 Absorp'on of Heat by Solids and Liquids If heat is put into a solid, the temperature rises. How much? Q = CΔT = C(T f T i ) Change of system energy with change of temperature Heat Transfer Heat Capacity [ J/ºC or cal/ºc ] - depends on the material Specific Heat (heat capacity mass) Q = cmδt = cm(t f T i ) Water : c water = 1 cal/g C = 1 Btu /lb F = 4190 J/kg C ~ independent of mass Molar Specific Heat (heat capacity # of particles) Q = c molar nδt 1 mol = elementary units

4 Heat Capacity (C) versus Specific Heat (c) C = cm Heat capacity specific heat

5 Absorp'on of Heat: Problem #18 42 A 20.0 g copper ring at C has an inner diameter of D= cm. An Al sphere at C has a diameter of d= cm. The sphere is placed on top of the ring and the two are allowed to come to thermal equilibrium, with no heat lost to the surroundings. The sphere just passes through the ring at the equilibrium temperature. What is the mass of the sphere??? First find the temperature where D=d ( ( )) ( ) ( ) D=D 0 1+ α Cu T f 0 ( ) ( ) = ( 1+ α Al ( T f 100) ) d f = d i 1+ α Al T f α Cu T f 0 T f = C Heat into ring Q=c Cu m r T f Heat from Sphere Q = c Al m s m s = c Al c Cum r T f ( T f T i ) ( T f T i )

6 Calorimetry If 0.20 kg of tea at 95 C is poured into a 0.15 kg glass cup initially at 25 C, what will be the final temperature T f of the mixture when equilibrium is reached, assuming no heat flows to the surroundings (c glass = 840 J/kg K & c water = 4190 J/kg K ) Heat lost = heat gained or Q=0

7 More Example: Hot Jogger C human body = 3500 J/(kg C) Q = c human _ body mδt In a half hour a 65-kg jogger can generate J of heat. If the heat were not removed by the various means possible, by how much will the body s temperature rise? In the relatively cool environmental temperature of 50 F, healthy marathon runners can have body temperatures as high as F. Weight lifters often have temperatures of 101 F during workouts in a warm gym. One runner who was still conscious is reported to have developed a temperature of F after finishing a marathon, but most people cannot tolerate temperatures that high. When your temperature rises above 102, your muscle often start to burn, when your temperature is over 104 you will usually become short of breath and when your temperature rises above 105, you will often have signs of brain distress, such as a headache, blurred vision, ringing in your ears, dizziness, nausea and passing out.

8 Heat Transforma'on Heat is transferred in or out of system, but temperature does NOT change: Change of phase Heat of Heat of Fusion Vaporization Q = ± L m ΔT = 0 Amount of heat transferred during phase change depends on L and mass (M) L F Heat of fusion Solid to liquid (heat is adsorbed : atomic bonds are broken L V Heat of Vaporization Liquid to gas (heat is adsorbed) L S Heat of Sublimation Solid to gas (heat is adsorbed) H 2 O L F = 79.5 cal/g = 6.01 kj/mol = 333 kj/kg L V = 539 cal/g = 40.7 kj/mol = 2256 kj/kg

9 Example: H 2 O H 2 O c ice = 0.53 cal/g K L F = 79.5 cal/g c water = 1.0 cal/g K L V = 539 cal/g 1 kg Making ice How much energy does a refrigerator have to remove from 1.5 kg of water at 20 C to make ice at -12 C. Will all the ice melt? At a party, a 0.5 kg chunk of ice at - 10 C is placed in a 3.0 kg of tea at 20 C. At what temperature and in what phase will the final mixture be?

10 H 2 O c ice = 0.53 cal/g K L F = 79.5 cal/g c water = 1.0 cal/g K L V = 539 cal/g Example 1 kg Making ice How much energy does a refrigerator have to remove from 1.5 kg of water at 20 C to make ice at -12 C. 1) Calculate the Q need to remove to cool water to freezing point 0 C Q = c water mδt = (4186 J kg C)(1.5kg)(0 20 C) = J 2) Calculate Q needed to fuse water into ice Q = L F m = ( J kg )(1.5kg) = J 3) Calculate Q needed to cool ice from 0 C to -12 C Q = c ice mδt = (2000 J kg C)(1.5kg)( 12 0) = J Q TOT = J J J = J

11 H 2 O c ice = 0.53 cal/g K L F = 79.5 cal/g c water = 1.0 cal/g K L V = 539 cal/g Example 1 kg Will all the ice melt? At a party, a 0.5 kg chunk of ice at - 10 C is placed in a 3.0 kg of tea at 20 C. At what temperature and in what phase will the final mixture be? 1) How much heat must be removed from tea(water) to reach 0 C? Q = c WATER m TEA ΔT = (4186 J )(3kg)( 20 C)= 2.5 kg C 105 J 2) How much heat must added to raise temperature of ice to 0 C? Q = c ICE m ICE ΔT = (2000 J )(0.5kg)(10 C) kg C =104 J 3) How much heat must added to melt ice? Q = L F m ICE = ( J )(0.5kg) =1.675 kg 105 J Less 4) What now? Ice warms to 0 C and melts at 0 C taking Q =10 4 J J = J than 5) What is final temperature? The melted ice warms and the tea cools down; the final temperature is Q cool tea =Q warm ice +Q melt ice +Q warm water c WATER m TEA (20 C T) = c WATER m ICE (T 0 C) J 60 3T = 0.5T 0 C (kg C) T = 5.0 C

12 Heat and Work If piston moves, the differential work done BY the gas is: dw by = F d s = pa ( )ds = p Ads ( ) = pdv In going from V i to V f, the total work done BY the gas is: V f W by = dw = pdv area under p-v graph During change in volume, the temperature and/or the pressure may change. V i How the system changes from state i to f determines how much work is done BY the system. WORK (and HEAT) IS (are) PATH DEPENDENT.

13 Path Dependence of Work W by = dw = pdv V f V i isobaric: W = pδv isochoric: W = 0 Volume increases, Pressure decreases: area > 0 -> W by > 0 (gas expands) Two step: Volume increases then Pressure decreases: area > 0 -> W by > 0 Here W by > 0, but less than before Work is NOT CONSERVATIVE: depends on path Volume decreases, Pressure increases: W by < 0 Work done by system is NEGATIVE (gas is compressed) NET WORK, W net, done by system (gas) during a complete cycle is shaded area. It can be pos., neg, or zero depending on path