Problem 1:
What is the hydrostatic force on the back of Grand Coulee Dam if the water in the reservoir is 150m deep and the width of the dam is 1200m.
Solution:
 | Concepts, principles, relations
that apply to the problem: The relationship between force and pressure The pressure at a point below the surface of a liquid at rest in a constant gravitational field depends only on the depth d of that point and the pressure at the surface. P = P0 + rgd. |
| Why do they apply? The water behind the dam at rest. |
| How do they apply? Let us denote the distance below the surface by y. The pressure on the side of the dam facing the water at depth y is P0 + rgy. Here P0 is the atmospheric pressure (101 kPa) and r is the density of water. The force on a surface element of width w and height dy at depth y is dF = (P0 + rgy)dA = (P0 + rgy)wdy. The total force on this side of the dam points outward and has magnitude  .(All quantities are measured in SI units.) |
| Details of the calculation: The total force on the side of the dam facing air is wP0150, inward. The net force on the dam is F = wrg1502/2, outward. With w = 1200m and r = 1000kg/m3, F = 1.32´1011N. |
Problem 2:
A Venturi tube may be used as a fluid flow meter. If the difference in pressure P1 - P2 = 21 kPa, find the fluid flow rate in m3/s given that the radius of the outlet tube is 1cm, the radius of the inlet tube is 2cm, and the fluid is gasoline (r = 700kg/m3).
Solution:
| Concepts, principles, relations
that apply to the problem: Bernoulli's equation: P + rgh + ½rv2 = constant. |
| Why do they apply? If we neglect friction, we have a conservative system. Bernoulli's equation is derived assuming the mechanical energy of the system is conserved. If a fluid or a gas, which is not being compressed, is flowing in a steady state, then the pressure depends on the speed of the fluid or the gas. The faster the fluid is flowing, the lower is the pressure at the same height. |
| How do they apply? P1 + rgh1 + ½rv12 = P2 + rgh2 + ½rv22; h is constant, so P1 + ½rv12 = P2 + ½rv22. P1 - P2 = ½rv22 - ½rv12. 21kPa = 350kg/m3(v22 - v12). From the equation of continuity we have Area 1 ´ v1 = Area 2 ´ v2. v1 = (A2/A1)v2. Inserting this into the above equation we have(1 - (A2/A1)2)v22) = (21000/350)(m/s)2. (A2/A1)2 = (1/4)2 = 1/16. v22 = (21000/350)(16/15)(m/s)2 = 64(m/s)2.. v2 = 8m/s. The fluid flow rate therefore is v2A2 = (8m/s)p(.01m)2 = 0.0025m3/s. |
| Details of
the calculations: None |
Problem 3:
A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. If the top piece is taken off and placed in the water, what happens to the water level in the tub?
Solution:
| Concepts, principles, relations that apply to the
problem: The buoyant force |
| Why do they apply? An object partially or wholly immersed in a gas or liquid is acted upon by an upward buoyant force B equal to the weight w of the gas or liquid it displaces. B = w |
|
How do they apply? The wood is floating, it is not accelerating. The weight of the displaced water must be equal to the weight of the wood. As long as both pieces of wood are floating, the amount of water displaced is independent of their configuration. The water level in the tub does not change when the configuration is changed. |
| Details of the calculation: None |
Problem 4:
A frog in a hemispherical pod finds that he just floats without sinking into a sea of blue-green ooze with density 1.35 g/cm3. If the pod has radius 6cm and negligible mass, what is the mass of the frog?
Solution:
| Concepts, principles, relations
that apply to the problem: The buoyant force |
| Why do they apply? When an object floats, the magnitude of the buoyant force is equal to the magnitude of its weight. |
| How do they apply? The magnitude of the buoyant force is equal to the the magnitude of the weight wliquid of the displaced liquid. wliquid = rliquidVg. The volume V of the displaced liquid is the volume of one half sphere, V = 2pr3/3 = 2p(6cm)3/3 = 452cm3. The magnitude of the weight of the object is wobject = mfrogg. (We are neglecting the weight of the air-filled pod.) rliquidVg = mfrogg. mfrog = 1.35(g/cm3)452cm3 = 610g. |
| Details of the calculation: None |
Problem 5:
A 5-liter vessel contains 0.125 moles of an ideal gas at a pressure of 1.5 atm. What is the average translational kinetic energy of a single molecule?
Solution:
| Concepts, principles, relations that apply to the
problem: Kinetic theory, the ideal gas law |
| Why do they apply? PV = nRT. The temperature is a direct measure of the average translational molecular kinetic energy. (1/2)m<v2> = (3/2)kBT. |
| How do they apply? PV = nRT yields T. T = PV/(nR) = (1.5´1.01´105 Pa)(5000 cm3 ´ 1 m3/106 cm3)/(0.125´8.31 J/K) = 729 K.  .The average translational kinetic energy of a single molecule is 1.51´10-20 J. |
| Details of the calculation: None |
Problem 6:
A tank having a volume of 0.1 m3 contains helium gas at 150 atm. How many balloons can the tank blow up, if each filled balloon is a sphere 0.3 m in diameter at an absolute pressure of 1.2 atm?
Solution:
| Concepts, principles, relations
that apply to the problem: The ideal gas law, (Boyle's law) |
| Why do they apply? At constant temperature P1V1 = P2V2. (Boyle's law) We are given P1, V1, and P2 and are asked to solve for V2. |
| How do they apply? P1 = 150 atm = 1.515´107 Pa. V1 = 0.1m3. P2 = 1.2 atm = 1.212´105 Pa. V2 = P1V1/P2 = 12.5 m3. Let n = number of balloons and Vb = the volume of each blown-up balloon. Vb = (4p/3)r3 = 1.414´10-2 m3. n = V2/Vb = 884. The tank can blow up 884 balloons. |
| Details of the calculation: None |
Problem 7:
A bar of gold is in thermal contact with a bar of silver of the same length
and area. One end of the compound bar is maintained at 80 oC and the opposite end is at 30 oC.
When the heat flow reaches steady state, find the temperature at the
junction.
The thermal conductivity of gold is 314 W/(moC), and the thermal
conductivity of silver is 427 W/(moC).
Solution:
| Concepts, principles, relations
that apply to the problem: Thermal conductivity The thermal conductivity k is defined through the equation  . |
| Why do they apply? When a steady state is reached, then the same amount of heat crosses any cross sectional area perpendicular to the bar per second. (Otherwise the energy would increase in certain regions, the temperature would increase there, the temperature gradient would change, and we would not have a steady state.) The amount of heat flowing is determined by the temperature difference and the thermal conductivity. |
| How do they apply? If the temperature at the junction is T, then in the gold we have DQ/Dt = (314 W/(moC))´A´(80 oC - T)/(L/2) and in the silver we have DQ/Dt = (427 W/(moC))´A´(T - 30 oC)/(L/2). We can therefore write (314 W/(moC))´A´(80 oC - T)/(L/2) = (427 W/(moC))´A´(T - 30oC)/(L/2). (314 W/(moC))´(80 oC - T) = (427W/(moC))´(T - 30 oC). (80 oC - T) = 1.36´(T - 30 oC). 120.8 oC = 2.36 T. T = 51.2 oC. |
| Details of the calculation: None |
Problem 8:
A mercury thermometer has a bulb of volume 0.100 cm3 at 10°C. The capillary tube above the bulb has a cross-sectional area of 0.012 mm2. The volume thermal expansion coefficient of mercury is 1.8*10-4 (°C)-1. How much will the mercury rise when the temperature rises by 20°C?
Solution:
| Concepts, principles, relations
that apply to the problem: The volume expansion coefficient The average volume expansion coefficient b is defined through DV = bVDT. |
| Why do they apply? The volume of the mercury will increase as the temperature rises and the mercury will rise in the capillary tube. |
| How do they apply? The increase in temperature is 20°C. The change in the volume of the mercury is DV = bVDT = (1.8*10-4 (°C )-1)(0.100 cm3)(20°C) = 3.6*10-4 = 0.36 mm3. Dh = DV/A = (0.36 mm3)/(0.012 mm2) = 30 mm. The mercury will rise 30 mm. |
| Details of the calculation: None |
Problem 9:
If 90 g of molten lead at 327.3 oC is poured into a 300 g casting form made of iron and initially at 20 oC, what is the final temperature of the system? Assume no energy is lost to the environment.
Solution:
| Concepts, principles, relations
that apply to the problem: Specific heat capacity: c = DQ/(mDT) The specific heat capacity c is the amount of energy it takes to raise the temperature of one kg of material by 1 degree Kelvin or Celsius. Latent heat of fusion: DQ = Lm The latent heat is the heat released or absorbed per unit mass by a system in a reversible isobaric-isothermal change of phase. |
| Why do they apply? Energy is conserved. The energy released by the lead is equal to the energy absorbed by the iron. |
| How do they apply? The melting point of lead is 327.3oC. Assume the final temperature of the system is T. Then the amount of energy released by the lead as it solidifies is DQ = mleadLlead = 0.09 kg´(2.45´104 J/kg) = 2205 J, and the amount of energy released as it cools is DQ = mleadcleadDT = 0.09 kg´(128 J/(kgoC))(327.3 oC - T) = (11.52 J/oC)(327.3 oC - T). This energy is absorbed by the iron. For the iron we therefore have 2205 J + (11.52 J/oC) (327.3oC - T) = mironcironDT = 0.3 kg´(448 J/(kgoC))(T - 20 oC). 5975.5 J - (11.52 J/oC)T = (134 J/oC)T - 2688 J. 8663.5 J = (145.52 J/oC)T. T = 59.5oC. |
| Details of the calculation: None |
Problem 10:
A liquid of unknown specific heat at a temperature of 20°C was mixed with water at 80°C in a well-insulated container. The final temperature was measured to be 50°C, and the combined mass of the two liquids was measured to be 240-g. In a second experiment with both liquids at the same initial temperatures, 20-g less of the liquid of unknown specific heat was poured into the same amount of water as before. This time the equilibrium temperature was found to be 52°C. Determine the specific heat of the liquid. The specific heat of water is 4187 J/Kg°C or 1 kcal/kg°C.
Solution:
| Concepts, principles, relations
that apply to the problem: Specific heat capacity: c = DQ/(mDT) The specific heat capacity c is the amount of energy it takes to raise the temperature of one kg of material by 1 degree Kelvin or Celsius. |
| Why do they apply? Energy is conserved. Knowing the specific heat capacity of the liquids we can calculate how it is shared between the liquids. |
| How do they apply? DQ = cmDT clml30 = cwmw30, clml = mw, ml + mw = 0.24 cl(ml –0.02)32 = cwmw28 |
| Details of the calculation: 3 equations, 3 unknowns ml = 0.16 mw = 0.08 cl = 0.5 kcal/(kg oC) |
Problem 11:
How much thermal energy is required to change a 40 g ice cube from a solid at -10 oC to steam at 110 oC?
Solution:
| Concepts, principles, relations
that apply to the problem: Specific heat capacity: c = DQ/(mDT) The specific heat capacity c is the amount of energy it takes to raise the temperature of one kg of material by 1 degree Kelvin or Celsius. Latent heat: DQ = Lm The latent heat is the heat released or absorbed per unit mass by a system in a reversible isobaric-isothermal change of phase. |
| Why do they apply? We are asked to find the amount of heat absorbed by a substance when its temperature increases and it changes phase twice. |
| How do they apply? To raise the temperature of the ice to 0 oC we need DQ = 0.04 kg´(0.49 kcal/(kgoC))10 oC = 0.196 kcal. To melt the ice we need DQ = 0.04 kg´80 kcal/kg = 3.2 kcal. To raise the temperature of the water to 100 oC we need DQ = 0.04 kg´(1 kcal/(kgoC))100 oC= 4 kcal. To boil the water we need DQ = 0.04 kg´540 kcal/kg = 21.6 kcal. To raise the temperature of the steam to 110oC we need DQ = 0.04kg´(0.48 kcal/(kgoC))10 oC = 0.192 kcal. The total thermal energy required is (0.196 + 3.2 + 4 + 21.6 + 0.192)kcal = 29.188 kcal. |
| Details of the calculation: None |
Problem 12:
A monatomic ideal gas, originally at a pressure P, volume V and temperature T, is compressed to one half of its initial volume.
| (a) If the compression is isothermal (i.e., at constant temperature), what is the final pressure? |
| (b) If the compression is isobaric (i.e., at constant pressure), what is the final temperature? |
| (c) What is the amount of heat supplied to the gas during the compression? |
Solution:
| Concepts, principles, relations
that apply to the problem: Ideal gas law: PV = NkT, PV/T = constant Energy conservation: increase in internal energy of a system = heat put into the system + work done on the system by its surroundings, or DU = DQ + DW. |
| Why do they apply? The system consist of a monatomic, ideal gas, for which the ideal gas law holds. |
| How do they apply? (a) T = constant: P1V1 = P2V2 = ½P2V1, P2 = 2P1. (b) P = constant: V1/T1 = V1/T2 = ½ V1/T2, T2 = ½ T1, (T is the absolute temperature). |
| Details of the calculation: (c) DU = DQ + DW. The work done on the system is -òPdV. Part (a): The temperature stays constant, the internal energy stays constant. The work done on the system is -òPdV = -NkTò(1/V)dV = NkT ln(Vi/Vf). The work done on the system is positive, (DW = NkTln(2) = PVln(2)), therefore DQ = - DW is negative. No heat is supplied to the gas, all the work done on the system increases the temperature of the surroundings. Part (b): The temperature decreases, the internal energy decreases. The work done on the system is positive (DW = PV/2), therefore DQ = -DW + DU = -PV/2 – 3PV/4 is negative. No heat is supplied to the gas, heat flows from the gas into its surroundings and all the work done on the system increases the temperature of the surroundings. (Note: For a monatomic ideal gas U = N(3/2)kT = (3/2)PV.) |
Problem 13:
One mole of an ideal gas is compressed at 60 oC isothermally from 5 atm to 20 atm.
|
(a) Find the work done. |
|
(b) Find the entropy change for the gas and interpret its algebraic sign |
Gas constant : 8.314 J/(mol K)
Solution:
| Concepts, principles, relations
that apply to the problem: Ideal gas law: PV = nRT, work done on the system: W = -òPdV Energy conservation: DU = DQ + DW Change in entropy:  ,
dS = dQ/TThe subscript r denotes a reversible path. |
| Why do they apply? Using the ideal gas law we can find the work done on the system. Using DU = DQ + DW we can then find the amount of heat transferred to or from the system. Knowing DQ allows us to calculate DS. |
| How do they apply? (a) The temperature is constant: -òPdV = -nRTò(1/V)dV = nRT ln(Vi/Vf) = nRT ln(Pf/Pi) = RT ln(4) W = (8.314 J/K) * (333 K) ln(4) = 3838 J |
| Details of
the calculations: (b) Energy conservation: increase in internal energy of a system = heat put into the system + work done on the system by its surroundings, or DU = DQ + DW. The temperature stays constant, the internal energy stays constant. The work done on the system is positive, therefore DQ = - DW is negative. DS = DQ/T = (-3838 J)/(333 K) = -11.5 J/K |
Problem 14:
One mole of an ideal gas does 3000 J of work
on its surroundings as it expands isothermally to a final pressure of 1 atm
and volume of 25 L. Determine
(a) the initial volume and
(b) the temperature of the gas.
Solution:
| Concepts, principles, relations
that apply to the problem: Ideal gas law: PV = nRT, work done on the system: W = -òPdV Energy conservation: DU = DQ + DW |
| Why do they apply? For an isothermal process W = nRT ln(Vi/Vf). We are given n, Pfinal, and Vfinal, and we are told that the temperature is constant. |
| How do they apply? (a) For an isothermal process the temperature is constant. Therefore PV = nRT = constant. PV = 101000 Pa´25´10-3 m3 = nRT W = nRT ln(Vi/Vf) for an isothermal process. W/(nRT) = -3000 J/(101000 Pa´25´10-3 m3) = -1.19 (The work done by the gas is -W.) Vf/Vi = exp(1.19) = 3.28 Vi = (25´10-3 m3)/3.28 = 7.62´10-3 m3 = 7.62 L (b) For an ideal gas PV = nRT. 101000Pa´25´10-3 m3 = (8.31 J/K)T. T = 303.85K. |
| Details of the calculation: None |
Problem 15:
An ice tray contains 500 g of water. Calculate the change in entropy of the water as it freezes completely and slowly at 0 oC.
Solution:
| Concepts, principles, relations
that apply to the problem: Change in entropy: ,
dS = dQ/TThe subscript r denotes a reversible path. |
| Why do they apply? We are asked calculate the change in entropy DS = DQ/T. While the water changes phase, the temperature stays constant. |
| How do they apply? DS = DQ/T. DQ = -mL, m = mass of water, L= latent heat of fusion = 333000 J/kg. DS = -(0.5 kg)(333000J /kg)/273 K = -610 J/K. |
| Details of the calculation: None |
Problem 16:
The surface of the Sun is approximately at 5700 K, and the temperature of the Earth's surface is approximately 290 K. What entropy changes occur when 1000 J of thermal energy is transferred from the Sun to the Earth?
Solution:
| Concepts, principles, relations
that apply to the problem: Change in entropy: ,
dS = dQ/TThe subscript r denotes a reversible path. |
| Why do they apply? We are asked calculate the change in entropy DS = DQ/T. |
| How do they apply? During the process the temperatures of the sun and the earth do not change appreciably. The change in the entropy of the sun is therefore DS = -1000 J/5700 K = -0,175 J/K. The change in the entropy of the earth is DS = 1000J/290 K = 3.448 J/K. The entropy of the sun-earth system increases by 3.27 J/K. |
| Details of the calculation: None |
Problem 17:
Determine the work that can be obtained from the one cycle of the ideal Carnot machine in which the working substance is a photon gas. The energy density of this gas is given by the u = σT4 and the pressure is P = (1/3)u.
Solution:
| Concepts, principles, relations
that apply to the problem: The second law of thermodynamics |
| Why do they apply? For an ideal, reversible engine Q1/T1 = Q2/T2. |
| How do they apply? For a Carnot engine we have Q1/T1 = Q2/T2, Wmax = (Qhigh - Qlow)reversible = Qhigh - QhighTlow/Thigh = Qhigh(1 - Tlow/Thigh). This is the same for all Carnot engines and does not depend on the working fluid. |
| Details of the calculation: None |
Problem 18:
A cylinder contains He gas. In its initial state a the cylinder
Volume is Va = 4.23*10-3 m3 and the pressure is
Pa = 1.19*10-5 Pa. The volume is isothermically reduced
until the system reaches a state b with Vb = 0.581*10-3
m3. This process is followed by an isobaric expansion which allows
the system to reach a state c with Vc = Va. The
cycle finishes with an isochoric or isometric (constant volume) process that
brings the system back to state a.
(a) Draw the cycle in a P-V diagram.
(b) Find the work done by the system going from b to c.
(c) Find the work done by the system in one cycle.
(d) Find the fractional change of the temperature of the system going from b
to c.
(e) When the system goes from c to a, does it absorb or release
heat? Find the amount of heat absorbed or released.
(f) When the system goes from b to c, does it absorb or release
heat? Find the amount of heat absorbed or released.
Solution:
| Concepts, principles, relations
that apply to the problem: Ideal gas law: PV = nRT, work done on the system: W = -òPdV Energy conservation: DU = DQ - DW increase in internal energy of a system = heat put into the system - work done by the system on its surroundings, |
| Why do they apply? An isothermal process occurs at constant temperature. P = nRT/V . Since the internal energy of a gas is only a function of its temperature, DU = 0 for an isothermal process. For the isothermal expansion of an ideal gas we have
W is negative if V2
< V1. Since DU
= 0, the heat transferred to the gas is
DQ
= W.
If the pressure of an ideal gas
is kept constant, then the temperature must increase as the gas expands.
(PV/T = constant.) Heat must be added during the expansion process. |
| How do they apply? (a)
(b) Wbc = Pb(Va-Vb)
= Pa(Va/Vb) (Va-Vb) =
3.16*10-7 J |
| Details of the calculation: None |
.
.