PHY Final Exam Problem 1 (20 points) A small ball of mass 2.00 kg is suspended by two wires, wire 1 and wire 2. If two wires make 90 o and wire 2 makes  =30 o with respect to the vertical, find the tension T 1 and T 2.  = 30 o 90 o T1T1 wire 1 T2T2 wire 2 Solution Let’s denote the length of wire 2 by L. Then the condition that the ball does not rotate about the pivot point of wire 2, Q, is: -L mg sin  + LT 1 =0 (5 points), namely, T 1 =mg sin  = (2.00 kg)(9.80 m/s 2 )sin30 o =9.80 N (5 points). The condition that the net force is zero in the horizontal direction gives: T 1 cos  - T 2 sin  = 0 (5 points), namely, T 2 =T 1 /tan30 o =17.0 N (5 points). (The condition that the net force is zero in the vertical direction gives: T 1 cos  + T 2 sin  = mg can be used too.) PQ mg

Problem 2 (10+10 points) An innovative musician is making a new music instrument using cylindrical glasses of height h=1 m with some water in them where the top (bottom) of each glass is open (closed). Right now he is adjusting the height of the water in one of the glasses so that sound of middle C note at frequency 256 Hz creates resonance. Assume that the speed of sound is 344 m/s. (a) What is the maximum height of water for the instrument to resonate with middle C note? (b) What is the next lowest frequency of sound that this glass instrument can resonate with if the water level is the same as in Part (a). open closed h water Solution (a)This instrument works as a stopped pipe. Therefore the frequencies of possible n-th harmonics are expressed by f n =nv/(4L n ) where n=1,3,5,…, v is the speed of sound, and L n is the height of air column. Since L n =nv/(4f), for f=256 Hz, L 1 becomes shortest at (344 m/s)/(4 x 256 1/s)=0.336 m when n=1. So the maximum water height for middle C note is m m=0.664 m. For a higher harmonics, the height of the air column is larger. (b) When the water level is adjusted as in Part (a), the fundamental frequency f 1 is 256 Hz. The next lowest frequency is when n=3, namely the third harmonics f 3 =3f 1 =3 x 256 Hz = 768 Hz. L

Problem 3 (20 points) Jane and John will host a big party. They are preparing iced tea for the party. Their plan is to have iced tea at 3.0 o C in an insulated container without ice left in it. They first made 15 liters of tea at 20 o C. Then they plan to put some ice at -5.0 o C into the container to cool the tea. How much of ice is needed to make this iced tea according to their plan? Treat the tea as water. If needed, use the following constants: the density of water  water =1000 kg/m 3, the latent heat of fusion of water L f =3.33 x 10 5 J/kg, specific heat of ice c ice =2090 J/(kg K) and the specific heat of water c water =4190 J/(kg K). Solution The mass of tea is m water =(15 L) x (1.0 kg/L)=15 kg (3 points). The amount of heat needed to melt the ice is Q melt =m ice x (3.33 x 10 5 J/kg) (3 points). To heat the ice from -5.0 o C to 0.0 o C the amount of heat needed is Q ice = m ice x c ice x [0.0 o C –(-5.0 o C)]=m ice x(10.5x10 3 J/kg) (3 points). The heat needed to raise the temperature of tea from the ice from 0.0 o C to 3.0 o C is Q ice-water =m ice x c water x [3.0-(0.0 o C)]=m ice x(12.6x10 3 J/kg) (3 points). The heat the tea acquires is Q water =m water x c water x [3.0 o C -(20.0 o C)]= (15 kg)x c water x(-17.0 o C)=1068 kJ (3 points). From Q melt +Q ice +Q ice-water +Q water =0 (2 points), m ice x(333x 10 3 J/kg) + m ice x(10.5x10 3 J/kg) + m ice x (12.6x10 3 J/kg) + (1068x10 3 J) = 0. So m ice x(356x10 3 J)=1068x10 3 J and therefore the amount of ice needed is m ice =3.0 kg (5 points).

Problem 4 ( points) Solution 1/s + 1/s’ = 1/f where s is the object distance, s’ is the image distance and f is the focal length. (a)f=10 cm and s = 25 cm, so s’=17 cm. The magnification m=-s’/s=16.7 cm/25 cm=-2/3. So the image is formed at 17 cm to the right of the lens, real and inverted. (b)f=10 cm and s=10 cm, so s’=+infinity. So the image is formed at infinity. (c)f=10 cm and s=7.5 cm, so s’=-30 cm. The magnification m=-s’/s=+4. So the image is formed at 30 cm to the left of the lens; virtual and upright. (d)f=10 cm and s=-20 cm, so s’=6.7 cm. The magnification m=-s’/s=+1/3. So the image is formed at 6.7 cm to the right of the lens; real and upright. (Note the object is a virtual image.) You are experimenting with a converging lens with a focal length of 10 cm. Find the image position (whether on the left or right of the lens and how far from the lens), magnification, and whether the image is upright or inverted, and real or virtual when an object is placed at (a) 25 cm, (b) 10 cm, (c) 7.5 cm and (d) -20 cm. If the image is formed at infinity, just stating that gets you 5 points. Assume that the incoming light rays come from the left of the lens toward the lens.

Problem 5 ( points) An experimental power plant at the Natural Energy Lab generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are 27 o C and 6 o C, respectively. (a) What is the maximum theoretical efficiency of this power plant? (b) If the power plant is to produce 210 kW of power, at what rate must heat be extracted from the warm water? (c) If the power plant is to produce 210 kW of power, at what rate must heat be absorbed by the cold water? (d) To raise the temperature of water whose mass is dm by , how much of heat dQ is needed? (For this question you may use the symbol c for specific heat of water). (e) The cold water that enters the plant leaves it at a temperature of 10 o C. What must be the flow rate of cold water through the system in kg/h? [Use, if needed, the fact that specific heat of water is c=4190 J/(kg K)]. (a) The maximum efficiency is realized by a Carnot engine. Therefore (b) (d) (c) (e) Solution

Problem 6 ( points) Three moles of an ideal gas are taken around the cycle acb shown in Figure. For this gas C p = 29.1 J/(mol K). Process ac is at constant pressure, process ba is at constant volume, and process cb is adiabatic. The temperature of the gas in states a,c, and b is T a =300 K, T c =492 K, and T b =600 K. The universal gas constant R is 8.31 J/(mol K). (a)For a process at constant pressure, when the temperature changes by  T, how much does the volume change  V? (b)For process ac, how much work is done (plug-in numbers)? (c)For an ideal gas, find the molar heat capacity at constant volume C V in terms of C p and the universal gas constant R (no numbers). (d) For process cb, how much is the change of internal energy,  U in terms of the number of moles n, C V, and the change of temperature  T (no numbers)? (e) For process cb, how much work is done (plug-in numbers)? (f)For process ba, how much work is done (plug-in numbers)? (g)Find the total work done for the cycle acba? p V O b ac (a)From the ideal gas equation, pV=nRT. So p  V=nR  T. (b)W ac =p  V=nR  T=(3.00 mol)[8.31 J/(mol K)](492 K – 300 K) =4.79x10 3 J. (c) C V =C p -R. (d)  U=nC V  T. (e) As Q cb =0 for an adiabatic process, W cb =Q cb -  U=-nC V  T=-n(C p -R)  T =-(3.00 mol)[29.1 J/(mol K)-8.31 J/(mol K)](600 K – 492 K)=-6.74x10 3 J. (f) For a process at constant volume, the work done is zero. So W ba =0.00 J. (g) The total work W tot =W ac +W cb +W ba =4.78x10 3 J – 6.74x10 3 J=-1.95x10 3 J. Solution