Final Exam

The final exam will take place Wednesday, April 30 at 8 am. The exam will consist of 1 to 12 problems and cover from Chapter 1 through Chapter 20. About half will be over the material from Ch 1-16 and about half over the new material, Ch 17-20. Below are a few sample exam problems for Ch 17-20.

You are responsible for all material covered in lecture and homework assignments. DO NOT ASSUME that a subject not covered in the sample exam below will not show up on Tuesday's exam! During the exam, you will be provided with an equation sheet. You may download the Equation sheet. You will be allowed to use only the equations on the sheet, or others that you derive from those on the sheet. Use only that when trying the problems below. Resist the temptation to look at the answers, as that will better prepare you for the exam.

1. You are designing a Hg thermometer that will use the full 1.0 m height of a cylindrical tube for a temperature range of 100^oC. If the reservoir bulb has volume 1 cm³, what should be the inner radius of the glass tube into which the Hg expands? For a 5 pt bonus, account for the expansion of the glass bulb. (b_Hg = 182 x 10^-6 /^oC and a_glass = 3.2 x 10^-6 /^oC )

   DV_Hg = b_HgVDT = inner volume of the tube = (pr²)L , -> r = 0.00761 cm without the glass correction

   For the bonus, the Hg must fill the glass tube and the increase in volume of the glass bulb:
   DV_Hg = b_HgVDT = (pr²)L + 3a_glassVDT, -> r = .00741 cm.

2. A Klingon ambassador offers Dr. Spock a 100 gm drink at 20 ^oC that is supposedly Saurian brandy (specific heat 1.5 cal/gm^oC). A suspicious Spock drops a 50 gm ice cube (0^oC) into the drink. There is 20 gm of ice left when equilibrium is reached. Assuming no heat losses to surroundings, is this Saurian brandy?

   Calculate c for the drink from, DQ₁ + DQ₂ + ... = 0. (100g)c(0-20) + (30gm)(80 cal/gm) = 0, so c = 1.2 cal/gm^oC. What effect would assuming that there is some heat loss have on this result? That is, does that assumption imply that this is definitely not Saurian brandy or could it be that it is Saurian Brandy and the heat loss could account for the low result? What other assumption(s) did you have to make?

3. 100 gm of a metal initially at 1000^oC is place in 200 gm of water at 20^oC, all in an insulating cup. Equilibrium is reached after 10 gm of water is boiled away as steam. Assuming no heat losses, what was the specific heat of the metal?

   Since equilibrium is reach after only part of the water boils away, then the final temperature of the metal and remaining 190 gm of water must be 100^oC. So our heat conservation equation is"

   (100 gm)(c _metal)(100-1000) + (200 gm)(1.0 cal/gm^oC)(100-20) + (10 gm)(540 cal/gm) = 0. Solve for c _metal = 0.24 cal/gm^oC

4. The walls of a 1.0 x 1.0 x 0.5 m freezer have thermal conductivity 0.008 J/s^oCm and are 5.0 cm thick. What minimum power is needed to maintain a -20^oC temperature inside if the outside temperature is 25^oC?

   ^DQ_Dt = kA^DT_Dx = (0.008)[2x(1.0x1.0) + 4x(1.0x0.5)]^[25-(-20)] / _0.05 = 29 Watts

    

5. How many molecules are there in a typical 1 L lungfull of air on a typical day at sea level? State what values you decide to use.

   PV = nRT -> (100,000Pa)(.001m³) = n(8.314)(~300K) -> n ~ .04 moles which is .04 x 6.02 x 10²³ = 2.4 x 10²² molecules.

   

6. One mole of O₂ gas undergoes a cycle consisting of an isotherm, an isobar, and finally an isochor. Find DQ, DU, and W for each of the three processes and for the entire cycle.

   1. For the isotherm, DU = 0, DQ = W = nRTln(V_f/V_i). We find T = PV/nR = 3x10⁵(0.025)/1(8.314) = 902, (or just insert PV for nRT), so W = 8200 J which is also DQ.

   2. For the isobar, W = P(DV) = 10⁵(0.025-0.075) = -5000 J, DQ = nC_pDT. The temperature starts at 902 and drops to 902/3 = 301 and C_p = C_V + R = (7/2)R for a diatomic gas, so DQ = - 17,500 J. And finally, we can get DU from nC_vDT or DU = DQ - W = -12,500 J.

   3. For the isochor, W = 0, and DQ = DU = nC_vDT = 1(5/2)R(902-301) = +12,500 J.

   Finally, for the whole cycle, simply add each segment. DU = 0 (always!), and the net heat exchange is the work out, DQ = W = 3200 J. Note the net work is positive, as to be expected for a clockwise cycle.

7. A steam engine operates between the temperature extremes of 120 ^oC and 45 ^oC.
   a) What is the maximum possible efficiency for this engine?
   b) If the engine generates 3,000 Watts of power, at what rate is heat wasted? (Assume the efficiency calculated in part a.)

   Eff = 1 - (273+45)/(273+120) = .19 = Work/Q_in --> Q_in = 3000 W/.19 = 15, 700 W and Q_out = Q_in - W = 12, 700 W. Note that we are really using power = Q/t in all cases. You can simply think of it as 3000 J of work every second, etc.