Physics Solution Manual for 1100 and 2101

Ssm reasoning and solution the rankine and fahrenheit

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Unformatted text preview: ospheric pressure. Technique B applies at pressures that are above normal, since the freezing point is then lower. Technique C works also, because the heat of vaporization is carried away by each kilogram of vapor that is removed. This heat comes from the remaining liquid, and since no heat can flow in from the surroundings to replenish this loss of heat, the remaining liquid will freeze. 16. 5.54 × 10−4 kg/s 17. (a) Relative humidity is defined by Equation 12.6 as the ratio (expressed as a percentage) of the partial pressure of water vapor in the air to the equilibrium vapor pressure of water at the existing temperature. The greater the temperature, the greater is the equilibrium vapor pressure of water. It is the equilibrium vapor pressure of water at the existing temperature, then, that determines the maximum amount of water that the air can contain. Warmer air can contain a greater maximum amount of water vapor than cooler air. Statement A is certainly true if both humidity values are quoted at the same temperature. Statement B could be true if the 30% humidity referred to a higher temperature than the 40% value. Likewise, statement C could be true if the 30% humidity referred to a higher temperature than the 40% value. 18. (b) The dew point is the temperature at which the existing partial pressure of water vapor in the air would be the equilibrium vapor pressure of water. The equilibrium vapor pressure of water always increases with increasing temperature, and only this answer is consistent with such a relationship. 628 TEMPERATURE AND HEAT CHAPTER 12 TEMPERATURE AND HEAT PROBLEMS ______________________________________________________________________________ 1. SSM REASONING AND SOLUTION 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 ______________________________________________________________________________ 2. REASONING a. According to the discussion in Section 12.1, the size of a Fahrenheit degree is smaller than that of a Celsius degree by a factor of 5 ; thus, 1 F° = 5 C° . This factor will be used to 9 9 find the temperature difference in Fahrenheit degrees. b. The size of one kelvin is identical to that of one Celsius degree (see Section 12.2), 1 K = 1 C°. Thus, the temperature difference, expressed in kelvins, is the same at that expressed in Celsius degrees. SOLUTION a. The difference in the two temperatures is 34 °C − 3 °C = 31 C°. This difference, expressed in Fahrenheit degrees, is 1 F° Temperature difference = 31 C° = ( 31 C° ) 5 9 C° = 56 F° b. Since 1 K = 1 C°, the temperature difference, expressed in kelvins, is 1K Temperature difference = 31 C° = ( 31 C° ) 1 C° = 31 K ____________________________________________________________________________________________ 3. REASONING a. The relationship between the Kelvin temperature T and the Celsius temperature TC is given by T = TC + 273.15 (Equation 12.1). Chapter 12 Problems 629 b. The relationship between the Kelvin temperature T and the Fahrenheit temperature TF can be obtained by following the procedure outlined in Examples 1 and 2 in the text. On the Kelvin scale the ice point is 273.15 K. Therefore, a Kelvin temperature T is T − 273.15 kelvins above the ice point. The size of the kelvin is larger than the size of a Fahrenheit degree by a factor of 9 . As a result a temperature that is T − 273.15 kelvins above the ice 5 point on the Kelvin scale is 9 (T − 273.15) Fº above the ice point on the Fahrenheit scale. 5 This amount must be added to the ice point of 32.0 ºF. The relationship between the Kelvin and Fahrenheit temperatures, then, is given by TF = 9 5 (T − 273.15) + 32.0 SOLUTION a. Solving Equation 12.1 for TC, we find that Day TC = T − 273.15 = 375 − 273.15 = 102 °C Night TC = T − 273.15 = 1.00 × 10 2 − 273.15 = −173 °C b. Using the equation developed in the REASONING, we find Day Night 4. TF = 9 5 (T − 273.15 ) + 32.0 = 9 ( 375 − 273.15) + 32.0 = 5 ( 215 °F ) TF = 9 (T − 273.15) + 32.0 = 9 1.00 ×102 − 273.15 + 32.0 = −2.80 ×102 °F 5 5 REASONING 1 A° is larger than 1 B°, because there are 90 A° between the ice and boiling points of water, while there are 110 B° between these points. Moreover, a given temperature on the A scale is hotter than the same reading on the B scale. For example, +20 °A is hotter than +20 °B, because +20 °A is 50 A° above the ice point of water while +20 °B is at the ice point. Thus, we expect +40 °A to correspond to more than +40 °B. SOLUTION a. Since there are 90.0 A° and 110.0 B° between the ice and boiling points of water, we have that 110.0 1 A° = B° = 1.22 B° 90.0 b. +40.0 °A is 70.0 A° above the ice point of water. On the B thermometer, this is 630 TEMPERATURE AND HEAT 1.22 B° = 85.4 B° above the ice point. 1...
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