Thurs Dec 3 Thermodynamics 1st Law 2 nd Law Heat Engines and Refrigerators Isobaric: W = PΔV Isochoric: W = 0 Isothermal: ΔU = 0 Adiabatic: Q = 0 Assign 13/14 Friday Final: Fri Dec 11 2:30PM Walter 145 2 Sheets handwritten 30% Thermal Physics 16 MC, 4 problems Exam Conflict? Notify me ASAP Study: Homework, PRS, Pre-class Examples in text Review problems Q/A Sessions: Thursday Dec 10 in Walter 245-2:30-4:30 PM - 7-9 PM
Given the PV diagram to the right and the information that ΔU AB is positive, what can you conclude about Q AB and W AB? (1) Q>0, W>0 (2) Q>0, W = 0 (3) Q>0, W<0 (4) Q=0, W>0 (5) Q=0, W=0 (6) Q=0, W<0 (7) Q<0, W>0 (8) Q<0, W=0 (9) Q<0, W<0 P A C B V From the graph, W is +. The only way ΔU can be + is if Q is + and greater than W. How the temperature changes?
Each drawing shows a hypothetical heat engine or a hypothetical heat pump and shows the corresponding heats and works. Only one is allowed in nature. Which is it? 1) 2) 3) 4) 5)
Refrigerator Compress refrigerant Heat it Expand Cool it Inside Q C Colder than in Refrig Expand Outside Q H Warmer than outside Motor Compress W
Refrigerators Coefficient of Performance (COP) For Carnot Refrigerator Q C /Q H = T C /T H :
Heat Engines, Refrigerators, Heat Pumps Heat Engine Natural flow of heat (hot to cold) can do work on surroundings Refrig or Heat Pump Unnatural flow of heat (cold to hot) requires work from surroundings Both cases: Q H = Q C + W What is related to what???? Think about Reservoirs, not the temp of something in the reservoir Which reservoir is hotter? Colder? Q C inside refrigerator; Q H Kitchen
Heating Your Kitchen How long would a 3.40 kw space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance = 3.29) does when it freezes 1.30 kg of water at 21.4 C into ice at 0 C? One way Another way Refrig Warms Kitchen Q Warms Kitchen Q C Q H Same Electrical Heater W Elect Energy = P*t Q H (Refrig) = Power (of elect heater) * time (elect heater run)
In cooling a warm leftovers, a refrigerator removes 60000J of heat from the food. The COP is 3.0. We eventually want to know the amount of heat exhausted into the room. First, though, which quantity does the 60000J represent? 1) Q C 2) Q H 3) W 60000J represents the amount of energy which must be removed from the cold reservoir.
In cooling warm leftovers, a refrigerator removes 60000J of heat from the food. If the COP is 3.0, what is the amount of heat exhausted to the room (moved into the hot reservoir)? 1) 20,000J 2) 60,000J 3) 80,000J 4) 180,000J 5) 240,000J COP = Q C /W 3.0 = 60000 J/W W = 20,000 J Q H = Q C +W = 60,000 J + 20,000 J
In cooling warm leftovers, a refrigerator removes 60000J of heat from the food. The COP is 3.0. If the power of the motor is 200W, what is the minimum time to cool the leftovers? 1) 100s 2) 200s 3) 300 s COP = Q C /W W = 20,000J P = Energy/Time Time = 100 s 200W = 20,000J/time
Heat Pump Coefficient of Performance (COP) a bit different Benefit: Energy from electrical lines leveraged to pull energy from outside. More efficient than electrical heater. Energy coming into house is W + Q C
Heat Engine, Heat Pump, and Refrigerators Summary Q H = W + Q C Device What we want What we buy Efficiency, e, or Coefficient of Performance ( COP ) Heat engine Carnot Heat pump Carnot Refrigerato r Carnot W net Q H 0 < e = W net / Q H = 1 - (Q C /Q H ) < 1 e = 1 - (T C /T H ) = (ΔΤ/Τ Η ) Q H W net COP = Q H /W net = 1 + [Q C /(Q H - Q C )] > 1 COP = 1/[1 - (T C /T H )] Q C W net COP = 1/[(T H /T C ) - 1] COP = Q C /W net (which may be more or less than 1)
A block of ice is held stationary against a wall as shown. What is the direction of frictional force between the ice and the wall? (Draw a free-body diagram) (1) left (2)right (3) up (4) down
A block of ice is held stationary against a wall as shown. What is the direction of normal force of the wall on the ice? (1) left (2)right (3) up (4) down
A block of ice is held stationary against a wall as shown. If the spring has a spring constant of 500 N/m, the spring is compressed 0.04 m, and the mass of the cube is 1.2kg, what is the magnitude of the normal force? (1) 1.2 N (2) 11.8 N (3) 20.0 N (4) 42.4 N (5) 500 N (6) 12500 N ΣF X = ma x F spring F normal = 0 F spring = F normal kx = F normal (500N/m)(0.04m) = F normal
A block of ice is held stationary against a wall as shown. If the spring has a spring constant of 500 N/m, the spring is compressed 0.04 m, and the mass of the cube is 1.2kg, what is the magnitude of the frictional force of the wall on the cube? (1) 0.12 N (2) 1.2 N (3) 9.8 N (4) 11.8 N (5) 20 N (6) 500 N ΣF y = ma y F friction F gravity = 0 F friction = F gravity F friction = m g F friction = (1.2kg)(9.8m/s 2 ) Normal Gravity (weight) Friction Spring
A baseball is hit and leaves the bat with a speed of 40m/s at an angle of 30 above the horizon. What is the speed of the ball at maximum height? (1) 0 m/s (2) 20 m/s (3) 35 m/s (4) 40 m/s At max height, v y = 0 m/s and v x = v x0 v x0 = (40m/s) cos30
A coin is tossed straight up into the air at a speed of 5 m/s. What is the maximum height of the coin as measured from the point it is released by the hand? (1) 0.26 m (2) 0.51 m (3) 1.3 m (4) 5.0 m (5) 9.8 m v y0 = 5 m/s v y = 0 m/s (at max height) a y = 9.8 m/s 2 Or conservation of energy: E final = E initial + W NC v 2 y = v 2 y0 + 2 a y y PE final = KE initial 0 = (5m/s) 2 + 2(-9.8m/s 2 ) y y = 1.28 m If path not straight, need to make some mention of energy. Can t just use kinematic formula even though it works out.
A father and his seven-year old daughter are facing each other on ice skates. With their hands, they push off against one another. Which one or more of the following statements is true? (1) Each experiences an acceleration that has a different magnitude (2) Each experiences an acceleration of the same magnitude (3) Each experiences a pushing force that has a different magnitude (4) Each experiences a pushing force of the same magnitude (5) 1 & 3 (6) 2 & 4 (7) 1 & 4 (8) 2 & 3 Same magnitude of force, but not the same mass (for most fathers & 7-year-old daughters!), so accelerations are different.
A stone is thrown from a cliff. Air resistance is negligible. As the stone falls, is it (1) in equilibrium (2) not in equilibrium The sum of the forces is not 0 (gravity is the only force) and the acceleration is not zero
A 2.0-kg block (A) is in contact with a spring (stiffness 400 N/m) that is compressed. When the spring is released, the block slides across the frictionless tabletop, reaching a maximum speed of 3.0 m/ s. Block A undergoes an elastic collision with an identical block (B). Block B slides across the tabletop and then encounters a 12-m long region where the frictional force on Block B is 1.5N. A B
What is conserved during the elastic collision between A and B? (1) combined momentum of Blocks A and B (2) combined kinetic energy of Blocks A and B (3) momentum of Block A only (4) kinetic energy of Block A only (5) Both (1) and (2) (6) Both (3) and (4) A B
What is the velocity of Block B after the collision? (1) 0 m/s (2) 1.2 m/s (3) 1.5 m/s (4) 2.0 m/s (5) 2.3 m/s (6) 3.0 m/s (7) 6.0 m/s (8) 9.4 m/s We saw that in a head-on elastic collision, if both objects have the same mass and one is at rest, the moving object just gives its velocity to the resting object
How far does Block B travel across the 12-m long frictional surface? 1) 0.46 m 2) 0.92 m 3) 4.0 m 4) 6.0 m 5) 8.0 m 6) 11.1 m 7) 11.5 m 8) all the way Method 1: Energy Block B has (1/2)(2 kg) (3 m/s) 2 = 9 J kinetic energy before hitting the friction, has zero when it stops After a distance d, work by frictional force is W= F f d W = KE f KE i F f d = 0 9 J (1.5 N)d = 9 J d = 6 m Method 2: Kinematics ΣF x = ma x 1.5 N = (2.0 kg)a x a x = 0.75 m/s 2 then v x 2 = v 0x 2 + 2a x x 0 = (3 m/s) 2 + 2( 0.75 m/s 2 )x x = 9/1.5 x = 6 m
By what distance was the spring originally compressed away from its equilibrium position? 1) 0.0075 m 2) 0.023 m 3) 0.045 m 4) 0.087 m 5) 0.15 m 6) 0.21 m 7) 0.46 m The energy of Block A s motion was originally elastic potential energy of the spring KE i + PE i = KE f + PE f 0 + (1/2)kx 2 = (1/2)mv 2 + 0 0.5 ( 400 N/m)x 2 = 9 J x 2 = (18 J)/(400 N/m) x = 0.21 m
The figure shows a skier at the bottom of a ski jump at a point where the track is circular with a radius of r. Assume that the positive direction is toward the center of the circle. Which equation represents the relationship between the forces and the velocity?
Prairie dog burrows are situated in such a way that the air blows at different speeds across the openings. This creates an air flow within the burrow that keeps the air fresh. What is the primary principle involved with this phenomena? (1) Bernoulli s Principle (2) Newton s Law (3) Pascal s Law (4) Archimede s Principle
If the air is blowing faster over the opening on the left and the openings are at roughly the same height, what is the direction of air flow inside the burrow? (1) From right to left (2) From left to right Higher velocity at left opening means lower pressure. Air within the burrow will flow from the higher pressure (the right) to the lower pressure (the left).