Thermodynamics Problem Set 1. 100 o C converted to both the Fahrenheit scale and the kelvin scale is which of the following? a. 238 o F, 373.15 K b. 88 o F, 273.15 K c. 238 o F, 273.15 K d. 212 o F, 373.15 K 2. The amount of heat necessary to raise a body one degree of temperature (K or o C) is called: a. one calorie b. the specific heat capacity c. the heat capacity d. the molar heat capacity 3. A guitarist tunes the steel strings of her electric guitar in the cool air-conditioned environment of the hotel room. The neck and body of the guitar is wood. The coefficient of linear expansion for steel is 1.1 X 10-5 ( o C) -1 and that for wood is 4.0 X 10-6 ( o C) -1. When she begins to play that eveninig under the hot stage lighting: a. Her guitar will sound slightly flat. b. Her guitar will sound slightly sharp. c. Her guitar will be in tune. d. impossible to determine from given information. 4. When a mercury thermometer at room temperature is placed over a flame, a. The column of mercury first dips slightly then begins to rise b. The column of mercury rises quickly at first, then slows down. c. The column of mercury rises at a steady pace from the beginning on. d. none of the above. 5. Which of the following volumes is closest to that occupied by one mole of an ideal gas at room temperature? a. 22.4 liters b. 24.4 liters c. 18.9 liters d. 1 liter 6. What is the minimum energy required to transform a 100 g piece of ice at -50 o C into steam? a. 77.0 kcal b. 15.0 kcal c. 74.5 kcal d. 77.5 kcal
7. As compared to the day (17 o C) what is the percentage increase in the rate of heat loss at night (7 o C) from a house maintained at 27 o C? a. 100% b. 50% c. 97% d. impossible to determine from the given information 8. If the temperature on the surface of the sun were doubled, the intensity of heat on the side of the planet Mercury facing the sun would be approximately: a. 2 times greater b. 4 times greater c. 8 times greater d. 16 times greater 9. During an adiabatic free expansion of a gas: I. The temperature remains constant. II. The entropy of the gas increases. III. The internal energy of the gas remains constant a. I b. I and III c. II and III d. I, II, and III 10. Using the key below, which of the following sequences represents the Carnot cycle? AA. adiabatic expansion BB. adiabatic compression CC. isothermal expansion DD. isothermal compression a. DD, AA, CC, BB b. CC, AA, DD, BB c. BB, DD, AA, CC d. more than one of the above 11. Which of the following statements is not true? a. The Carnot cycle is reversible. b. The entropy of the universe has increased after a complete Carnot cycle. c. The Carnot cycle is not 100% efficient. d. all of the above are true. 12. What is the maximum efficiency of an engine operating between 177 o C and 27 o C? a. 33% b. 85% c. 50% d. 15%
The following passage pertains to the questions on this page (13-17). A car is traveling down a road in Spring. Although the temperature is above freezing, there are still snow drifts beside the road. The car loses control and plows into one of the snow drifts, completely submerging its tires. Before the wreck, the car's four tires contained a total of 30.3 L of air (assume the average molecular weight to be 16 g mol -1, specific heat = 0.25 cal g -1 o C -1 ) at 8 X 10 6 Pa and 30 o C. 13. What is the mass of the air in the tires? 14. a. 1.6 X 10 4 g b. 1.0 X 10 7 g c. 1.6 X 10 8 g d. 1.6 X 10 3 g Which of the following most closely approximates the behavior of the tires after they are submerged in the snow? a. isothermal compression b. adiabatic compression c. isobaric compression d. more than one of the above 15. Which of the following statements is true? 16. I. As heat flows from a tire into the snow, the entropy of the tire increases. II. As heat flows from a tire into the snow, the entropy of the snow increases. III. The magnitude of the entropy change in the snow is greater than the magnitude of entropy change in the tire. a. I b. I and III c. II and III d. I, II, and III Assuming that the air in the tires (1600 g) behaves like an ideal gas, and assuming constant pressure, approximately how much snow is melted by heat flow from the air in the tires alone? a. 15 g b. 25 g c. 45 g d. impossible to determine from given information 17. The entire process of heat flowing from the tires into the snow bank could be reversed if: a. if a machine such as a heat pump were constructed to extract heat from the snow bank and transmit it into the tires at the rate of heat loss from the tires b. the heat flow occured at a slow enough rate c. the second law of thermodynamics precludes the reversibility of the process d. if we divide the heat flow rate per unit area by the temperature gradient
18. If an air-conditioner that runs at 25% efficiency under certain internal and external temperature conditions were turned around in the window to act as a heat pump. What would its COP equal? a. 2.5 b. 4.0 c..50 d. Such a device would violate the second law of thermodynamics. 19. If the fire below a pot of boiling water on a gas stove were doubled, so that heat flow into the water were also doubled: a. The rate of entropy change associated with the heat flow into the boiling water would be sixteen times greater. b. The rate of entropy change associated with the heat flow into the boiling water would be doubled. c. The rate of entropy change associated with the heat flow into the boiling water would remain the same. d. The water would get hotter.
1. D The conversion from the celcius to the Fahrenheit scale is as follows: T f = 9 5 T c + 32 7. A Heat flow through a body is proportional to the temperature difference across the body: ΔQ Δt = kaδt Δx 2. C 3. A 4. A Here is the conversion from celcius to Kelvin (Kelvin, being the true thermodynamic temperature, gets the plain symbol T): T = T C + 273.15 The heat capacity is the quantity of heat required to raise the temperature of a body (a stone, the chair, etc.) one degree Kelvin. The specific heat capacity is the quantity required to raise the temperature of one gram of a substance one degree Kelvin. The molar heat capacity is the same for one mole of a substance. The strings, having a greater coefficient of linear expansion than the neck of the quitar, will expand more at the higher temperature than the neck. Thus they will lose a bit of tension, sounding slightly flat. The first thing that occurs upon heating is the expansion of the metal bulb, causing the mercury to drop at first. 8. D 9. D 10.* D 11. B The rate of heat radiation by a body is proportional to the fourth power of the temperature: P = σ AεT 4 Picture a membrane separating a gas within a canister from an area of vacuum. When the membrane is broken, the gas will undergo an adiabatic free expansion. The expansion will require no work and involve no heat flow. However, the entropy of the gas will increase. If you don't understand why, try to reverse the process. All three are possible Carnot cycles. Choice B represents the most familiar version, in which heat flows from the hot to the cold sink. The other two are the Carnot cycle in reverse (the action of a heat pump or refrigerator) in which we move the heat from the cold to the hot sink. That the Carnot cycle is reversible, is another way of saying that its action does not increase the entropy of the universe. 5. B 6.* C One mole of an ideal gas at STP occupies 22.4 liters (answer A). However, att room temperature, approximately 298 K, one mole of an ideal gas occupies 24.4 liters. The necessary heat equals the amount energy required to bring the water to 0 o C (The specific heat capacity of ice is.50 cal/g o C (or K), at which the point the heat of fusion must be added, which does not raise the temperature. Then we must bring the water from 0 o C to 100 o C (1 cal/g o C), after which we allow the heat of vaporization to flow into the system. 12. A The maximum efficiency (Carnot efficiency) is given by: (use Kelvin!) ε = W Q h = 1 Q c Q h = 1 T c T h 13.* D To find how many grams we have, first we must compute the number of moles of gas present in a system defined by given values of the state functions of our gas (P, V, T), using the ideal gas law (the equation of state of an ideal gas): PV = nrt
14. C 15. C 16.* B This is one of those MCAT questions where none of the answers is perfect. We're then left with choosing the best answer. We know that A and B are both obviously incorrect because heat will definately be flowing from the tires into the snow until thermal equilibrium is achieved. Pressure does definately go down with decreasing internal energy at constant volume. However, the weight of the car combined with the elasticity of the tires will ameliorate this effect somewhat by decreasing the volume. The entropy change of heat flow into an object at constant temperature will satisfactorily approximates the situation: ΔS = ΔQ r T The object into which heat flows receives a positive entropy change. The snow is at a lower temperature than the tire. This results in its positive entropy change being of greater magnitude (absolute value) than the negative entropy change of the tire. This fact corresponds to the entropy of the universe increasing, and thus the irreversibility of the process, when heat flows from the higher temperature to the lower temperature body. Remeber the molar heat capacity of an ideal gas: C v = 3 2 R 17. C 18. B 19. B Because the external temperature is above freezing, we can assume that the snow is at 0 o K. Therefore, only the heat of fusion is required to bring about melting: If each stage of a process is an equilibrium state (as in the Carnot cycle) then the universe gains no entropy thereby, and the transformation is reversible. This is definately not the case here. Without the addition of energy, all spontaneous processes are irreversible. When we run a Carnot cycle backwards, what we are basically doing is pulling heat from the cold sink through isothermal expansion, and then expelling the heat through isothermal compression on the hot sink. There is more heat expelled by an amount of added work. The ratio of the amount of heat expelled to the work we put in is called the coefficient of performance of the heat pump: COP = Q h W = T h T h T c Temperature is constant in boiling water. The entropy change to the water as heat flows in is given by: ΔS = ΔQ r T Converting our mass to moles, then employing a form of the above with calories as our units, we can compute the heat flow from the tires due to the temperature drop: Don't forget, however, that volume is decreasing. This means that the environment will be performing work on our system, adding energy to the tires. This energy, as heat, must also eventually flow out as temperature falls in the tires. At constant pressure, we know by Charles' law that, when the temperature ( o K) decreases by 10%, so also does the volume. The original volume was about 30 L, so it decreases by 3 L (3 X 10-3 m 3 ):