advertisement

PHYS 1401 General College Physics I (43172 - MTWR 5:30-7:50 PM) Home Work 12: (pages 366-369) 1, 5, 9, 13, 15, 29, 35, 43, 57, 67 1. The difference between these two averages, expressed in Fahrenheit degrees, is 98.6 F 98.2 F 0.4 F Since 1 C° is equal to 9 F°, we can make the following conversion 5 1 C (0.4 F) 0.2 C (9/5) F 5. The temperature of –273.15 °C is 273.15 Celsius degrees below the ice point of 0 °C. This number of Celsius degrees corresponds to (9/5) F (273.15 C) 491.67 F 1 C Subtracting 491.67 F° from the ice point of 32.00 °F on the Fahrenheit scale gives a Fahrenheit temperature of –459.67 F 9. Using the value for the coefficient of thermal expansion of steel given in Table 12.1, we find that the linear expansion of the aircraft carrier is L = L0 T = (12 10–6 C–1)(370 m)(21 °C 2.0 °C) 0.084 m (12.2) 13. The value for the coefficient of thermal expansion of steel is given in Table 12.1. The relation, L = L0T, written in terms of the diameter d of the rod, is d 0.0026 cm T 110 C (12.2) 6 d0 12 10 C 1 2.0026 cm 15. Assuming that the rod expands linearly with heat, we first calculate the quantity L/ T using the data given in the problem. L 8.47 10–4 m 1.13 10 –5 m/C T 100.0 C – 25.0 C Therefore, when the rod is cooled from 25.0 °C, it will shrink by L (1.13 10 –5 m/C ) T (1.13 10 –5 m/C ) (0.00 C – 25.0 C) –2.82 10 –4 m 29. According to Equation 12.3, V VT . Taking the coefficient of volumetric expansion for water from Table 12.1, we find that Vair V T 3.7 10–3 (C)–1 air 0 air 18 Vwater water V0 T water 207 10 –6 (C)–1 35. In order to keep the water from expanding as its temperature increases from 15 to 25 °C, the atmospheric pressure must be increased to compress the water as it tries to expand. The magnitude of the pressure change P needed to compress a substance by an amount V is, according to Equation 10.20, P B(V / V0 ) . The ratio V / V0 is, according to Equation 12.3, V / V0 T . Combining these two equations yields P B T SOLUTION Taking the value for the coefficient of volumetric expansion for water from Table 12.1, we find that the change in atmospheric pressure that is required to keep the water from expanding is 1 P (2.2 109 N/m 2 ) 207 106 C (25 C 15 C) 1 atm 4.6 106 Pa 45 atm 5 1.0110 Pa 43. The metabolic processes occurring in the person’s body produce the heat that is added to the water. As a result, the temperature of the water increases. The heat Q that must be supplied to increase the temperature of a substance of mass m by an amount T is given by Equation 12.4 as Q = cmT, where c is the specific heat capacity. The increase T in temperature is the final higher temperature Tf minus the initial lower temperature T0. Hence, we will solve Equation 12.4 for the desired final temperature. SOLUTION From Equation 12.4, we have Q cmT cm Tf T0 Solving for the final temperature, noting that the heat is Q = (3.0 × 105 J/h)(0.50 h) and taking the specific heat capacity of water from Table 12.2, we obtain 3.0 105 J/h 0.50 h Q Tf T0 21.00 C 21.03 C cm 4186 J/ kg C 1.2 103 kg 57. The heat required is Q = mLf + cmT, where m = V. See Table 12.2 for the specific heat c and Table 12.3 for the latent heat Lf. Thus, Q = VLf + cVT Q = (917 kg/m3)(4.50 10–4 m)(1.25 m2){33.5 104 J/kg + [2.00 103 J/(kgC°)](12.0 C°)} 1.85 10 5 J 67. From inspection of the graph that accompanies this problem, a pressure of 3.5 106 Pa corresponds to a temperature of 0 °C. Therefore, liquid carbon dioxide exists in equilibrium with its vapor phase at 0 C when the vapor pressure is 3.5 106 Pa .