ch19hw - PART 3 THERMODYNAMICS CHAPTER 19 TEMPERATURE AND...

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Unformatted text preview: PART 3 THERMODYNAMICS CHAPTER 19 TEMPERATURE AND HEAT Section 19-1: Macroscopic and Microscopic Descriptions Problem 1. The macroscopic state of a carton capable of holding a half-dozen eggs is specified by giving the number of eggs in the carton. The microscopic state is specified by telling where each egg is in the carton. How many microscopic states correspond to the macroscopic state of a full carton? Solution If we number the eggs (so that they are distinguishable), we could put the first egg in any one of six places, the second in any one of the remaining five places, etc. The total number of microscopic states is 6 5 4 3 2 1 = 6! = 720 . (If the eggs are indistinguishable, there is only one microscopic state for a full carton.) Section 19-3: Measuring Temperature Problem 2. A Canadian meteorologist predicts an overnight low of- 15 C . How would a U.S. meteorologist express that same prediction? Solution We assume that the U.S. meteorologist predicts the same temperature, but expresses it on the Fahrenheit scale (Equation 19-3): T F = 9 5 (- 15) + 32 = 5 F . Problem 3. Normal room temperature is 68 F . What is this in Celsius? Solution Equation 19-3, solved for the Celsius temperature, gives T C = 5 9 ( T F- 32) = 5(68- 32) / 9 = 20 C . Problem 4. The outdoor temperature rises by 10 C . What is the rise in Fahrenheit? Solution Temperature differences on the Fahrenheit and Celsius scales are related by T F = (9 / 5) T C , so ( 9 5 ) (10 C ) = 18 F . (Note that a temperature difference and a temperature reading are not the same, even though both are specified in the same units. The notation C versus C is an attempt to clarify this distinction, but is not universally accepted or consistently applied.) Problem 5. At what temperature do the Fahrenheit and Celsius scales coincide? Solution In Equation 19-3, T F and T C are numerically equal when T F = ( 9 5 ) T C + 32 = T C , or T C =- ( 5 4 ) (32) =- 40 = T F . Problem 6. Give equations for Rankine temperature in terms of (a) kelvins and (b) F . Solution (a) Temperature differences on the Rankine and Kelvin scales are related in the same way as those on the Fahrenheit and Celsius scales (see Problem 4), i.e., T R = ( 9 5 ) T. Both the Rankine and Kelvin scales are absolute scales and have the same zero point, therefore T R = ( 9 5 ) T. (b) If we use Equations 19-2 and 3, T R = ( 9 5 ) ( T C + 273 . 15) = T F- 32 + ( 9 5 ) (273 . 15) = T F + 459 . 67 . Problem 7. The normal boiling point of nitrogen is 77.3 K. Express this in Celsius and Fahrenheit. Solution Equations 19-2 and 3 give T C = 77 . 3- 273 . 15 - 196 C , and T F = ( 9 5 ) (- 196) + 32 =- 321 F . 2 CHAPTER 19 Problem 8. A sick childs temperature reads 39.1 on a Celsius thermometer. Whats the childs temperature on the Fahrenheit scale?...
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This note was uploaded on 09/15/2009 for the course PHY 557 taught by Professor Rijssenbeek during the Spring '08 term at Adelphi.

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ch19hw - PART 3 THERMODYNAMICS CHAPTER 19 TEMPERATURE AND...

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