Physics Solution Manual for 1100 and 2101

# Since the air conditioner is a carnot air conditioner

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Unformatted text preview: cmwater ( 85.0 °C ) − mice Lf c ( mice + mwater ) ( ) 4186 J/ ( kg ⋅ C° ) ( 6.00 kg )( 85.0 °C ) − ( 3.00 kg ) 33.5 ×104 J/kg = = 30.0 °C 4186 J/ ( kg ⋅ C° ) ( 3.00 kg + 6.00 kg ) We have taken the specific heat capacity of 4186 J/ ( kg ⋅ C° ) for water from Table 12.2 and the latent heat of 33.5 ×104 J/kg from Table 12.3. This temperature is equivalent to Tf = (273 + 30.0 °C) = 303 K. The change ∆Suniverse in the entropy of the universe is the sum of three contributions: [Contribution 1] T ∆S water = mwater c ln f T i 303 K = (6.00 kg)[4186 J/ ( kg ⋅ C° ) ] ln = − 4190 J/K 358 K Chapter 15 Problems 815 where Ti = 273 + 85.0 °C = 358 K. [Contribution 2] ∆Sice mLf Q = = = T T ( 3.00 kg ) ( 33.5 ×104 J/kg ) 273 K = +3680 J/K [Contribution 3] T ∆Sice water = mice c ln f T i 303 K = (3.00 kg)[4186 J/ ( kg ⋅ C° ) ] ln = + 1310 J/K 273 K The change in the entropy of the universe is ∆Suniverse = ∆Swater + ∆Sice + ∆Sice water = +8.0 × 10 2 J/K b. The entropy of the universe increases , because the mixing process is irreversible. 80. REASONING When an amount Q of heat flows spontaneously from the hot reservoir at temperature TH = 373 K to the cold reservoir at temperature TC = 273 K, the amount of energy rendered unavailable for work is Wunavailable = T0∆Suniverse (15.19) In Equation 15.19, T0 is the Kelvin temperature of the lowest-temperature reservoir (T0 = 173 K) that is available, and ∆Suniverse is the change in entropy of the universe as a result of the heat transfer. This entropy change is the sum of the entropy change ∆SH of the hot reservoir and the entropy change ∆SC of the cold reservoir: ∆Suniverse = ∆SH + ∆SC (1) Q To find the entropy changes of the hot and cold reservoirs, we employ ∆S = T R (Equation 15.18), imagining reversible processes where an amount of heat Q is transferred from the hot reservoir to the cold reservoir. The entropy changes given by Equation 15.18 will be the same as for the irreversible process that occurs as heat Q flows from the hot reservoir, through the copper rod, and into the cold reservoir, because entropy is a function of state. Equation 15.18, then, yields ∆S H = −Q TH and ∆S C = +Q TC (2) 816 THERMODYNAMICS The hot reservoir loses heat, and the cold reservoir gains heat, as heat flows from the hot reservoir to the cold reservoir. The explicit algebraic signs (− and +) in Equation (2) reflect these facts. Because the heat flow from the hot reservoir to the cold reservoir occurs as a process of thermal conduction, the heat Q that flows through the rod in a time t is found from Q= ( Here, k = 390 J s ⋅ m ⋅ Co ) ( kA ∆ T ) t (13.1) L is the thermal conductivity of copper (see Table 13.1 in the text), A and L are the cross-sectional area and length of the rod, respectively, and ∆T is the difference in temperature between the ends of the rod: ∆T = TH − TC. SOLUTION Substituting Equation (1) into Equation 15.19 yields Wunavailable = T0∆Suniverse = T0 ( ∆SH + ∆SC ) (3) Substituting Equations (2) into Equation (3) and simplifying, we find that −Q Q T T Wunavailable = T0 ( ∆S H + ∆SC ) = T0 + = Q − 0 + 0 TH TC TH TC (4) Lastly, substituting Equation 13.1 into Equation (4), we obtain a final expression for the amount of energy that becomes unavailable for doing work in this process: T T ( kA ∆T ) t T0 T0 Wunavailable = Q − 0 + 0 = + − TH TC L TH TC (5) 60 s Expressed in seconds, the elapsed time is t = 2.0 min = 120 s . In addition, the 1 min temperature difference ∆T = 373 K − 273 K = 1.00×102 K is equivalent to 1.00×102 C°, because one kelvin is equivalent to one Celsius degree. Equation (5), then, yields the desired result: ( Wunavailable ( )( ) )( ) 390 J s ⋅ m ⋅ Co 9.4 × 10−4 m 2 1.00 × 102 Co (120 s ) 173 K 173 K = + − 0.35 m 373 K 273 K = 2100 J Chapter 15 Problems 817 81. REASONING a. According to the discussion in Section 15.11, the change ∆Suniverse in the entropy of the universe is the sum of the change in entropy ∆SH of the hot reservoir and the change in entropy ∆SC of the cold reservoir, so ∆Suniverse = ∆SH + ∆SC. The change in entropy of each reservoir is given by Equation 15.18 as ∆S = (Q/T)R . The engine is irreversible, so we must imagine a process in which the heat Q is added to or removed from the reservoirs reversibly. T is the Kelvin temperature of a reservoir. Since heat is lost from the hot reservoir, the change in entropy is negative: ∆SH = − QH /TH. Since heat is gained by the cold reservoir, the change in entropy is positive: ∆SC = + QC /TC. The change in entropy of the universe is ∆S universe = ∆S H + ∆SC = − QH TH + QC TC b. The magnitude W of the work done by any engine depends on its efficiency e and input heat QH via W = e QH (Equation 15.11). For a reversible engine, the efficiency is related to the Kelvin temperatures of its hot and cold reservoirs by e = 1 − (TC/TH), Equation 15.15. Combining these two rel...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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