Chapter 20 Solutions

Transcription

1 Chapter 20 Solutions 20.1 Taking m = 1.00 kg, we have U g = mgh = (1.00 kg)(9.80 m/s 2 )(50.0 m) = 490 J But U g = Q = mc T = (1.00 kg)(4186 J/kg C) T = 490 J so T = C T f = T i + T = ( ) C Goal Solution G: Water has a high specific heat, so the difference in water temperature between the top and bottom of the falls is probably less than 1 C. (Besides, if the difference was significantly large, we might have heard about this phenomenon at some point.) O: The temperature change can be found from the potential energy that is converted to thermal energy. The final temperature is this change added to the initial temperature of the water. A: The change in potential energy is U = mgy and the change in internal energy is E int = mc T so mgy = mc T Therefore, T = gy c = (9.80 m/s2 )(50.0 m) J/kg C = C T f = T 1 + T = 10.0 C C = 10.1 C L: The water temperature rose less than 1 C as expected; however, the final temperature might be less than we calculated since this solution does not account for cooling of the water due to evaporation as it falls. It is interesting to note that the change in temperature is independent of the amount of water The container is thermally insulated, so no heat flows: Q = 0 and E int = Q W output = 0 W output = +W input = 2mgh. For convenience of calculation, we imagine setting the water on a stove and putting in this same amount of heat. Then we would have 2mgh = E int = Q = m water c T. T = 2mg h m water c = kg(9.80 m/s2 )3.00 m (0.200 kg)(4186 J/kg C ) = 88.2 J 837 J/C = C m m Thermal insulator

2 2 Chapter 20 Solutions 20.3 Q = mc silver T 1.23 kj = (0.525 kg)c silver (10.0 C) c silver = kj/kg C *20.4 From Q = mc T, we find T = Q mc = 1200 J ( kg)(387 J/kg C) = 62.0 C Thus, the final temperature is 87.0 C 20.5 Q cold = Q hot (mc T) water = (mc T) iron (20.0 kg)(4186 J/kg C)(T f 25.0 C) = (1.50 kg)(448 J/kg C)(T f 600 C) T f = 29.6 C Goal Solution G: Even though the horseshoe is much hotter than the water, the mass of the water is significantly greater, so we might expect the water temperature to rise less than 10 C. O: The heat lost by the iron will be gained by the water, and from this heat transfer, the change in water temperature can be found. A: Q iron = Q water or (mc T) iron = (mc T) water (1.50 kg)(448 J/kg C)(T 600 C) = (20.0 kg)(4186 J/kg C)(T 25.0 C) T = 29.6 C L: The temperature only rose about 5 C, so our answer seems reasonable. The specific heat of the water is about 10 times greater than the iron, so this effect also reduces the change in water temperature. In this problem, we assumed that a negligible amount of water boiled away, but in reality, the final temperature of the water would be less than what we calculated since some of the heat energy would be used to vaporize the water. *20.6 Let us find the energy transferred in one minute. Q = [m cup c cup + m water c water ] T Q = (0.200 kg) J 900 kg C + (0.800 kg) J 4186 ( 1.50 C ) = 5290 J kg C

3 Chapter 20 Solutions 3 If this much energy is removed from the system each minute, the rate of removal is Q = t = 5290 J 60.0 s = 88.2 J s = 88.2 W *20.7 (a) Q cold = Q hot (m w c w + m c c c )(T f T c ) = m Cu c Cu (T f T Cu ) m unk c unk (T f T unk ) where w is for water, c the calorimeter, Cu the copper sample, and unk the unknown. (250 g) 100 g C cal ( ) C g C = (50.0 g) g C unk( ) C cal = ( g C)c unk or c unk = cal/g C The material of the sample is beryllium m = ( m 3 )(1000 kg/m 3 ) (a) Q = mc T = Pt = ( kg)(4186 J/kg C)(1.00 C) Q = J = Pt t = J J/s = s = 53.1 yr 20.9 (a) (f)(mgh) = mc T (0.600)( kg)(9.80 m/s)(50.0 m) J/cal = (3.00 g)( cal/g C )( T) T = C; T = 25.8 C No Both the change in potential energy and the heat absorbed are proportional to the mass; hence, the mass cancels in the energy relation.

4 4 Chapter 20 Solutions Q cold = Q hot m Al c Al (T f T c ) + m c c w (T f T c ) = m h c w (T f T h ) (m Al c Al + m c w w )T f (m Al c Al + m c c w )T c = m h c w T f + m h c w T h (m Al c Al + m c c w + m h c w )T f = (m Al c Al + m c c w )T c + m h c w T h T f = (m Alc Al + m c c w )T c + m h c w T h (m Al c Al + m c c w + m h c w ) The rate of collection of heat = = (550 W/m 2 )(6.00 m 2 ) = 3300 W. The amount of heat required to raise the temperature of 1000 kg of water by 40.0 C is: Q = mc T = (1000 kg)(4186 J/kg C )(40.0 C) = J Thus, t = J or t = J 3300 W = 50.7 ks = 14.1 h *20.12 The heat needed is the sum of the following terms: Q needed = (heat to reach melting point) + (heat to melt) + (heat to reach boiling point) + (heat to vaporize) + (heat to reach 110 C) Thus, we have Q needed = kg[(2090 J/kg C)(10.0 C) + ( J/kg) + (4186 J/kg C)(100 C) + ( J/kg) + (2010 J/kg C)(10.0 C)] Q needed = J The bullet will not melt all the ice, so its final temperature is 0 C. Then 1 2 mv2 + mc T bullet = m w L f where m w is the meltwater mass m w = 0.500( kg)(240 m/s) 2 + ( kg)(128 J/kg C )(30.0 C ) J/kg

5 Chapter 20 Solutions 5 m w = 86.4 J J J/kg = g

6 6 Chapter 20 Solutions Goal Solution G: The amount of ice that melts is probably small, maybe only a few grams based on the size, speed, and initial temperature of the bullet. O: We will assume that all of the initial kinetic and excess internal energy of the bullet goes into internal energy to melt the ice, the mass of which can be found from the latent heat of fusion. A: At thermal equilibrium, the energy lost by the bullet equals the energy gained by the ice: K bullet + Q bullet = Q ice 1 2 m bv 2 + m b c lead T = m ice L f 1 2 ( kg)(240 m/s) 2 + ( kg)(128 J/kg C)(30.0 C) = m ice ( J/kg) 86.4 J J m ice = J/kg = kg = g L: The amount of ice that melted is less than a gram, which agrees with our prediction. It appears that most of the energy used to melt the ice comes from the kinetic energy of the bullet (88%), while the excess internal energy of the bullet only contributes 12% to melt the ice. Small chips of ice probably fly off when the bullet makes impact. So some of the energy is transferred to their kinetic energy, so in reality, the amount of ice that would melt should be less than what we calculated. If the block of ice were colder than 0 C (as is often the case), then the melted ice would refreeze. *20.14 (a) Q 1 = heat to melt all the ice = ( kg)( J/kg) = J Q 2 = (heat to raise temp of ice to 100 C) = ( kg)(4186 J/kg C)(100 C) = J Thus, the total heat to melt ice and raise temp to 100 C = J Q 3 = heat available as steam condenses = ( kg)( J/kg) = J Thus, we see that Q 3 > Q 1, but Q 3 < Q 1 + Q 2.

7 Chapter 20 Solutions 7 Therefore, all the ice melts but T f < 100 C. Let us now find T f Q cold = Q hot ( kg)( J/kg) + ( kg)(4186 J/kg C)(T f 0 C) = ( kg)( J/kg) ( kg)(4186 J/kg C)(T f 100 C) From which, T f = 40.4 C Q 1 = heat to melt all ice = J [See part (a)] Q 2 = heat given up as steam condenses = (10 3 kg)( J/kg) = J Q 3 = heat given up as condensed steam cools to 0 C = (10 3 kg)(4186 J/kg C)(100 C) = 419 J Note that Q 2 + Q 3 < Q 1. Therefore, the final temperature will be 0 C with some ice remaining. Let us find the mass of ice which must melt to condense the steam and cool the condensate to 0 C. ml f = Q 2 + Q 3 = J Thus, m = 3 J J/kg = kg = 8.04 g Therefore, there is 42.0 g of ice left over Q = m Cu c Cu T = m N2 (L vap ) N2 (1.00 kg) cal g C ( ) C = m 48.0 cal g m = kg *20.16 Q cold = Q hot [m w c w + m c c c ](T f T i ) = m s [ L v + c w (T f 100)] (0.250 kg) J 4186 kg C + ( kg) J 387 (50.0 C 20.0 C) kg C = m s J kg + J 4186 (50.0 C 100 C) kg C

8 8 Chapter 20 Solutions m s = 4 J = kg = 12.9 g steam J/kg

9 Chapter 20 Solutions (a) Since the heat required to melt 250 g of ice at 0 C exceeds the heat required to cool 600 g of water from 18 C to 0 C, the final temperature of the system (water + ice) must be 0 C. Let m represent the mass of ice that melts before the system reaches equilibrium at 0 C. Q cold = Q hot ml f = m w c w (0 C T i ) m( J/kg) = (0.600 kg)(4186 J/kg C)(0 C 18.0 C) m = 136 g, so the ice remaining = 250 g 136 g = 114 g The original kinetic energy all becomes thermal energy: 1 2 mv mv2 = ( kg)(500 m/s) 2 = 1.25 kj Raising the temperature to the melting point requires Q = mc T = kg(128 J/kg C )(327 C 20.0 C) = 393 J Since 1250 J > 393 J, the lead starts to melt. Melting it all requires Q = ml = ( kg) ( J/kg) = 245 J Since 1250 J > J, it all melts. If we assume liquid lead has the same specific heat as solid lead, the final temperature is given by J = 393 J J kg(128 J/kg C )(T f 327 C) T f = 805 C Q cold = Q hot m Fe c Fe ( T) Fe = m Pb [ L f + c T)] Pb (0.300 kg) J 448 kg C (T f 20.0 C) = kg J kg + J 128 kg C (T f C) and T = 59.4 C

10 10 Chapter 20 Solutions (a) W = P dv = P V = (1.50 atm)(4.00 m 3 ) = J W = P dv = P V = (1.50 atm)( )m 3 = J W if = f i P dv P f The work done by the gas is just the area under the curve P = αv 2 between V i and V f. P = αv α 2 i W if = f i αv 2 dv = 1 3 α (V3 f V 3 i ) O 1.00 m m 3 V V f = 2V i = 2(1.00 m 3 ) = 2.00 m 3 W if = atm Pa 3 m atm *20.22 (a) W = PdV [(2.00 m3 ) 3 (1.00 m 3 ) 3 ] = 1.18 MJ P(Pa) = ( Pa)( )m i + ( Pa)( )m 3 + ( Pa)( )m a) f W i f = MJ V(m 3 ) W f i = 12.0 MJ During the heating process P = (P i /V i )V. (a) W = i f 3V PdV = i(pi /V i ) VdV V i W = (P i /V i ) V2 2 3V i V i = P i 2V i (9V 2 i V 2 i) = 4P i V i PV = nrt [(P i /V i )V]V = nrt T = (P i /nrv i )V 2 Temperature must be proportional to the square of volume, rising to nine times its original value.

11 Chapter 20 Solutions W = i f P dv = P i f dv = PVf PV i W = nrt f nrt i W n = R( T) = 20.0 J (8.315 J/mol K)(100 K) = mol m = nm = ( mol) 4.00 g mol = g W = P V = P nr P (T f T i ) = nr T = 0.200(8.315)(280) = 466 J W = i f P dv = P i f dv = P( V) = nr( T) = nr(t2 T 1 ) (a) Along IAF, W = (4.00 atm)(2.00 liter) = 8.00 L atm = 810 J Along IF, W = 5.00 L atm = 506 J (c) Along IBF, W = 2.00 L atm = 203 J P(atm) 4 I A B F V(liters) (a) W = P V = (0.800 atm)( 7.00 L) = 567 J E int = Q W = 400 J J = 167 J E int = Q W Q = E int + W = 500 J 220 J = 720 J Positive heat is transferred from the system.

12 12 Chapter 20 Solutions (a) Q = W = Area of triangle = 1 2 (4.00 m3 )(6.00 kpa) = 12.0 kj P(kPa) 8 B Q = W = 12.0 kj A C * Q W E int BC 0 (Q = E int since W BC = 0) CA ( E int < 0 and W < 0, so Q < 0) AB (W > 0, E int > 0 since E int < 0 for B C A; so Q > 0) V(m 3 ) W BC = P B (V C V B ) = 3.00 atm( ) m 3 = 94.2 kj E int = Q W P(atm) 3.0 B C E int,c E int,b = ( ) kj E int,c E int,b = 5.79 kj Since T is constant, E int,d E int,c = 0 A D V(m 3 ) W DA = P D (V A V D ) = 1.00 atm( ) m 3 = 101 kj E int,a E int,d = 150 kj ( 101 kj) = 48.7 kj Now, E int,b E int,a = [(E int,c E int,b ) + (E int,d E int,c ) + (E int,a E int,d )] E int,b E int,a = [5.79 kj kj] = kj (a) E int = Q P V = 12.5 kj (2.50 kpa)( )m 3 = 7.50 kj V 1 = V 2 ; T T 1 T 2 = V 2 T 2 V 1 = (300 K) = 900 K (a) W = nrt ln V f V i = P f V f ln V f V i so V i = V f exp W P f V f V i = (0.0250) exp 3000 (0.0250)( ) = m 3

13 Chapter 20 Solutions 13 T f = P f V f nr = ( N/m 2 )( m 3 ) (1.00 mol)(8.315 J/K mol) = 305 K

14 14 Chapter 20 Solutions W = P V = P(V s V w ) = P(nRT) P P (18.0 g) (1.00 g/cm 3 )(10 6 cm 3 /m 3 ) W = (1.00 mol)(8.315 J/K mol)(373 K) ( N/m 2 ) 18.0 g 10 6 g/m 3 W = 3.10 kj Q = ml v = ( kg)( J/kg) = 40.7 kj E int = Q W = 37.6 kj (a) W = P V = P [3αV T] = ( N m 2 )(3)( )(C ) kg kg/m 3 W = 48.6 mj Q = cm T = (900 J/kg C)(1.00 kg)(18.0 C) = 16.2 kj (c) E int = Q W = 16.2 kj 48.6 mj = 16.2 kj (a) P i V i = P f V f = nrt = (2.00 mol)(8.315 J/K mol)(300 K) = J V i = nrt P i V f = nrt P f = J atm = J 1.20 atm = 1 3 V i = m 3 W = P dv = nrt ln V f V i = ( ) ln 1 3 = 5.48 kj (c) E int = 0 = Q W Q = 5.48 kj

15 Chapter 20 Solutions The condensing and cooling water loses heat ml v + mc T = kg [ J/kg + (4186 J/kg C )90.0 C ] Q = J From the First Law, Q = E int + W = 0 + nrt ln(v f /V i ) J = 10.0 mol (8.315 J/mol K)(273 K) ln(20.0 L/V i ) 2.09 = ln(20.0 L/V i ) V i = 2.47 L W = W AB + W BC + W CD + W DA W = B P dv + C P dv + D P dv + A P dv A B C D P 2 P B C W = nrt 1 B dv A V + P C 2 dv B + nrt 2 D dv C V + P 1 A dv D P 1 D A V V 1 V 2 W = nrt 1 ln V B V 1 + P 2 (V C V B ) + nrt 2 ln V 2 + P V C 1 (V A V D ) Now P 1 V A = P 2 V B and P 2 V C = P 1 V D, so only the logarithmic terms do not cancel out: Also, V B V 1 = P 1 P 2 and V 2 V C = P 2 P 1 W = nrt 1 ln P 1 P 2 + nrt 2 ln P 2 P 1 = nrt 1 ln P 2 P 1 + nrt 2 ln P 2 P 1 = nr(t 2 T 1 ) ln P 2 P 1 Moreover P 1 V 2 = nrt 2 and P 1 V 1 = nrt 1 W = P 1 (V 2 V 1 ) ln P 2 P 1

16 16 Chapter 20 Solutions E int,abc = E int,ac (conservation of energy) (a) E int,abc = Q ABC W ABC (First Law) P A B Q ABC = 800 J J = 1300 J W CD = P C V CD, V AB = V CD, and P A = 5P C Then, W CD = 1 5 P A V AB = 1 5 W AB = 100 J D C ( means that work is done on the system) V (c) W CDA = W CD so that Q CA = E int,ca + W CDA = 800 J 100 J = 900 J ( means that heat must be removed from the system) (d) E int,cd = E int,cda E int,da = 800 J 500 J = 1300 J and Q CD = E int,cd + W CD = 1300 J 100 J = 1400 J = ka(t 2 T 1 ) L = (0.200 cal/s cm C )(20.0 cm)(5000 cm)(180 C ) 1.50 cm = cal/s = ka T L k = L A T = (10.0 W)( m) (1.20 m 2 )(15.0 C) = W/m C = ka T L = (0.800 W/m C)(3.00 m2 )(25.0 C) m = W = 10.0 kw = A T i = L i k i (6.00 m 2 )(50.0 C) 2( m) W/m C m W/m C = 1.34 kw

17 Chapter 20 Solutions In the steady state condition, Au = Ag so that k Au A Au T x Au = k Ag A Ag T x Ag 80.0 C Au Ag 30.0 C In this case A Au = A Ag, x Au = x Ag, T Au = (80.0 T) Insulation and T Ag = (T 30.0) where T is the temperature of the junction. Therefore, k Au (80.0 T) = k Ag (T 30.0) and T = 51.2 C Two rods: = (k 1 A 1 + k 2 A 2 ) T L = (k 1 A 1 + k 2 A 2 ) (T h T c ) L T c L 1 2 T h In general: Insulation = ( k i A i ) T L = ( k ia i ) (T h T c ) L From Table 20.4, (a) R = ft 2 F h/btu The insulating glass in the table must have sheets of glass less than 1/8 inch thick. So we estimate the R-value of a inch air space as (0.250/3.50) times that of the thicker air space. Then for the double glazing R b = ft2 F h 3.50 Btu = 1.85 ft2 F h Btu (c) Since A and (T 2 T 1 ) are constants, heat flow is reduced by a factor of = = σaet 4 = ( W/m 2 K 4 ) [4π( m) 2 ](0.965)(5800 K) 4 = W

18 18 Chapter 20 Solutions *20.49 Suppose the pizza is 70 cm in diameter and l = 2.0 cm thick, sizzling at 100 C. It cannot lose heat by conduction or convection. It radiates according to = σaet 4. Here, A is its surface area, A = 2πr 2 + 2πrl = 2π(0.35 m) 2 + 2π(0.35 m)(0.02 m) = 0.81 m 2 Suppose it is dark in the infrared, with emissivity about 0.8. Then = ( W/m 2 K 4 )(0.81 m 2 )(0.80)(373 K) 4 = 710 W ~ 10 3 W If the density of the pizza is half that of water, its mass is m = ρv = ρπr 2 l = (500 kg/m 3 )π(0.35 m) 2 (0.02 m) 4 kg Suppose its specific heat is c = 0.6 cal/g C. The drop in temperature of the pizza is described by: Q = mc(t f T i ) = dq = mc dt f 0 dt f = mc = 710 J/s (4 kg)( J/kg C ) = 0.07 C /s ~ 10 1 K/s *20.50 = σaet W = ( W/m 2 K 4 )( m 2 )(0.950)T 4 T = [ K 4 ] 1/4 = K *20.51 We suppose the earth below is an insulator. The square meter must radiate in the infrared as much energy as it absorbs, = σaet 4. Assuming that e = 1.00 for blackbody blacktop: 1000 W = ( W/m 2 K 4 )(1.00 m 2 )(1.00)T 4 T = ( K 4 ) 1/4 = 364 K (You can cook an egg on it.) *20.52 The sphere of radius R absorbs sunlight over the area of its day hemisphere, projected as a flat circle perpendicular to the light: πr 2. It radiates in all directions, over area 4πR 2. Then, in steady state, in = out e(1340 W/m 2 )πr 2 = eσ(4πr 2 )T 4 The emissivity e, the radius R, and π all cancel. Therefore, T = 1340 W/m 2 1/4 = 277 K = 4 C 4( W/m 2 K 4 )

19 Chapter 20 Solutions 19 * K = C is the boiling point of nitrogen. It gains no heat to warm as a liquid, but gains heat to vaporize: Q = ml v = (0.100 kg)( J/kg) = J The water first loses heat by cooling. Before it starts to freeze, it can lose Q = mc T = (0.200 kg)(4186 J/kg C )(5.00 C ) = J The remaining ( )J = J that is removed from the water can freeze a mass x of water: Q = ml f J = x( J/kg) x = kg = 47.7 g of water can be frozen *20.54 The energy required to melt 1.00 kg of snow is Q = (1.00 kg)( J/kg) = J The force of friction is f = µn = µmg = (0.200)(75.0 kg)(9.80 m/s 2 ) = 147 N Therefore, the work done is W = fs = (147 N)s = J and s = m (a) Before conduction has time to become important, the heat energy lost by the rod equals the heat energy gained by the helium. Therefore, (ml v ) He = (mc T) Al or (ρvl v ) He = (ρvc T) Al so V He = (ρvc T) Al (ρl v ) He (2.70 g/cm V He = 3 )(62.5 cm 3 )(0.210 cal/g C)(295.8 C) (0.125 g/cm 3 )( J/kg)(1.00 cal/4.186 J)(1.00 kg/1000 g) = cm 3 = 16.8 liters The rate at which energy is supplied to the rod in order to maintain constant temperatures is given by = ka dt dx = (31.0 J/s cm K)(2.50 cm2 ) K 25.0 cm = 917 W This power supplied to the helium will produce a "boil-off" rate of

20 20 Chapter 20 Solutions ρlv = 917 W (0.125 g/cm 3 )( J/kg)(10 3 kg/g) = 351 cm3 s = L s

21 Chapter 20 Solutions 21 Goal Solution G: Demonstrations with liquid nitrogen give us some indication of the phenomenon described. Since the rod is much hotter than the liquid helium and of significant size (almost 2 cm in diameter), a substantial volume (maybe as much as a liter) of helium will boil off before thermal equilibrium is reached. Likewise, since aluminum conducts rather well, a significant amount of helium will continue to boil off as long as the upper end of the rod is maintained at 300 K. O: Until thermal equilibrium is reached, the excess heat energy of the rod will be used to vaporize the liquid helium, which is already at its boiling point (so there is no change in the temperature of the helium). A: As you solve this problem, be careful not to confuse L (the conduction length of the rod) with L v (the heat of vaporization of the helium). (a) Before heat conduction has time to become important, we suppose the heat energy lost by half the rod equals the heat energy gained by the helium. Therefore, (ml v ) He = (mc T) Al or (ρvl v ) He = (ρvc T) Al so that v He = (ρvc T) Al (ρl v ) He = (2.7 g/cm3 )(62.5 cm 3 )(0.21 cal/g C)(295.8 C) (0.125 g/cm 3 )(4.99 cal/g) and v He = cm 3 = 16.8 liters Heat energy will be conducted along the rod at a rate of dq = P = ka T L. During any time interval, this will boil a mass of helium according to Q = ml v or dq = dm L v Combining these two equations gives us the "boil-off" rate: dm = ka T L L v Set the conduction length L = 25 cm, and use k = 31 J/s cm K = 7.41 cal/s cm K: dm = (7.41 cal/s cm K)(2.5 cm2 )(295.8 K) (25 cm)(4.99 cal/g) = 43.9 g/s or dm = 43.9 g/s g/cm 3 = 351 cm3 /s = liter/s L: The volume of helium boiled off initially is much more than expected. If our calculations are correct, that sure is a lot of liquid helium that is wasted! Since liquid helium is much more expensive than liquid nitrogen, most low-temperature equipment is designed to avoid unnecessary loss of liquid helium by surrounding the liquid with a container of liquid nitrogen.

22 22 Chapter 20 Solutions *20.56 (a) The heat thus far gained by the copper equals the heat loss by the silver. Your down parka is an excellent insulator. Q cold = Q hot or m Cu c Cu (T f T i ) Cu = m Ag c Ag (T f T i ) Ag (9.00 g)(387 J/kg C )(16.0 C) = (14.0 g)(234 J/kg C )(T f 30.0 C) Ag (T f 30.0 C) Ag = 17.0 C so T f,ag = 13.0 C For heat flow: m Ag c Ag dt Ag = m Cu c Cu dt Cu dt dt = m Cuc Cu m Ag Ag c Ag dt Cu (9.00 g)(387 J/kg C ) = ( C /s) (14.0 g)(234 J/kg C ) = C /s (negative sign decreasing temperature) Ag Q = cm T; m = ρv; dq = ρc T dv c = dq/ ρ T(dV/) = (30.0 J/s) (0.780 g/cm 2 )(4.80 C)(4.00 cm 3 /s) = 2.00 J/g C c = 2.00 kj/kg C Q = mc T = (ρv)c T Thus, when a constant temperature difference T is maintained, the rate of adding heat to the liquid is = dq = ρ dv c T = ρrc T and the specific heat of the liquid is c = ρr T Call the initial pressure P 1. In the constant volume process 1 2 the work is zero. P 1 V 1 = nrt 1 P 2 V 2 = nrt 2 so P 2 V 2 P 1 V 1 = T 2 T 1 T 2 = 300 K 1 4 (1) = 75.0 K

23 Chapter 20 Solutions 23 Now in 2 3 W = 3 P dv = P2 (V 3 V 2 ) = P 3 V 3 P 2 V 2 2 W = nrt 3 nrt 2 = (1.00 mol) (8.315 J/mol K)(300 K 75.0 K) W = 1.87 kj *20.60 (a) Work done by the gas is the area under the PV curve P V i W = P i 2 V i = P i V i 2 b c In this case the area under the curve is W = P dv. Since the process is isothermal, a PV = P i V i = 4P V i i 4 = nrt i and V W = V i V i /4 dv V (P iv i ) = P i V i ln V i /4 V i = P i V i ln 4 = 1.39P i V i (c) The area under the curve is 0 and W = (a) The work done during each step of the cycle equals the area under that segment of the PV curve 3P i P B C W = W DA + W AB + W BC + W CD W = P i (V i 3V i ) P i (3V i V i ) + 0 = 4P i V i P i A D The initial and final values of T for the system are equal. V i 3V i V Therefore, E int = 0, and Q = W = 4P i V i (c) W = 4P i V i = 4nRT i = 4(1.00)(8.315)(273) = 9.08 kj (a) Fv = (50.0 N)(40.0 m/s) = 2000 W Energy received by each object is (1000 W)(10 s) = 10 4 J = 2389 cal. The specific heat of iron is cal/g C, so the heat capacity of each object is = cal/ C. T = 2389 cal cal/ C = 4.47 C

24 24 Chapter 20 Solutions The power incident on the solar collector is i = IA = (600 W/m 2 )π (0.300 m) 2 = 170 W For a 40.0% reflector, the collected power is c = 67.9 W The total energy required to increase the temperature of the water to the boiling point and to evaporate it is Q = cm T + ml V = (0.500 kg)[(4186 J/kg C )(80.0 C ) J/kg] Q = J The time required is t = Q c = J 67.9 W = 5.31 h From Q = ml v the rate of boiling is described by = Q t = L vm t m t = L v Model the water vapor as an ideal gas P 0 V = nrt = m M RT P 0 V t = m RT t M P 0 Av = RT L v M v = RT ML v P 0 A = v = 3.76 m/s (1000 W)(8.315 J/mol K)(373 K) ( kg/mol)( J/kg)( N/m 2 )( m 2 )

25 Chapter 20 Solutions 25 *20.65 To vaporize water requires an addition of J/kg of energy, while water gives up J/kg as it freezes. The heat to vaporize part of the water must come from the heat of fusion as some water freezes. Thus, if x kilograms of water freezes while a mass of (1.00 kg x) is vaporized, ( J/kg)x = ( J/kg)(1.00 kg x) or x = 6.79 kg 6.79x This yields, 7.79x = 6.79 kg, and x = kg = 872 g of water freezes Energy goes in at a constant rate. For the period from 50.0 min to 60.0 min, Q = mc T T ( C) 0 C) (10.0 min) = (10 kg + m i )(4186 J/kg C )(2.00 C (10.0 min) = 83.7 kj + (8.37 kj/kg)m i (1) For the period from 0 to 50.0 min, Q = m i L f (50.0 min) = m i ( J/kg) Substitute = m i ( J/kg)/50.0 min into Equation (1) to find t (min) m i ( J/kg)/5.00 = 83.7 kj + (8.37 kj/kg)m i 83.7 kj m i = = 1.44 kg ( ) kj/kg *20.67 (a) The block starts with K i = 1 2 mv2 i = 1 2 (1.60 kg)(2.50 m/s) 2 = 5.00 J. All this becomes extra internal energy in the ice, melting some according to "Q" = m ice L f. Thus, the mass of ice that melts is m ice = "Q" L f = K i 5.00 J = L f J/kg = kg = 15.0 mg For the block: Q = 0 (no heat can flow since there is no temperature difference), W = J, E int = 0 (no temperature change), and K = 5.00 J. For the ice, Q = 0, W = 5.00 J, E int = J, and K = 0. Again, K i = 5.00 J and m ice = 15.0 mg. For the block of ice: Q = 0, E int = J, K = 5.00 J, so W = 0. For the copper, nothing happens: Q = E int = K = W = 0.

26 26 Chapter 20 Solutions (c) Again, K i = 5.00 J. Both blocks must rise equally in temperature: "Q" = mc T = "Q" mc = 5.00 J 2(1.60 kg)(387 J/kg C ) = C At any instant, the two blocks are at the same temperature, so for both Q = 0. For the moving block: K = 5.00 J and E int = J, so W = J. For the stationary block: K = 0, E int = J, so W = 2.50 J A = A end walls + A ends of attic + A side walls + A roof A = 2(8.00 m 5.00 m) (4.00 m) (4.00 m tan 37.0 ) + 2(10.0 m 5.00 m) m 4.00 m cos 37.0 A = 304 m 2 = ka T L = ( kw/m C)(304 m 2 )(25.0 C) m = 17.4 kw = 4.15 kcal s Thus, the heat lost per day = (4.15 kcal/s)( s) = kcal/day. The gas needed to replace this loss = kcal/day 9300 kcal/m 3 = 38.6 m 3 /day LρAdx = ka T x Lρ x dx = k T 0 t Lρ x = k T t ( J/kg)(917 kg/m 3 ) ( m) 2 ( m) 2 2 = W 2.00 m C (10.0 C ) t t = s = 10.2 h For a cylindrical shell of radius r, height L, and thickness dr, Equation 20.14, dq = ka dt dx, becomes dq = k(2πrl) dt dr

27 Chapter 20 Solutions 27 Under equilibrium conditions, dq is constant; therefore, dt = dq 1 2πkL dr r and T b T a = dq 1 2πkL ln b a but T a > T b. Therefore, dq = 2πkL(T a T b ) ln (b/a) From problem 70, the rate of heat flow through the wall is T b T a dq = 2πkL(T a T b ) ln (b/a) r dq = 2π( cal/s cm C )(3500 cm)(60.0 C ) ln(256 cm/250 cm) L dq = cal/s = 9.32 kw b a This is the rate of heat loss from the plane, and consequently the rate at which energy must be supplied in order to maintain a constant temperature Q cold = Q hot or Q Al = (Q water + Q calo ) m Al c Al (T f T i ) Al = (m w c w + m c c c )(T f T i ) w (0.200 kg)c Al (+39.3 C ) = (0.400 kg) J 4186 kg C + ( kg) J 630 ( 3.70 C ) kg C c Al = J = 800 J/kg C 7.86 kg C

28 28 Chapter 20 Solutions