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# chap 16 - TEMPERATURE AND HEAT 16 TF = EXERCISES Section...

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16.1 TEMPERATURE AND HEAT E XERCISES Section 16.1 Heat, Temperature, and Thermodynamic Equilibrium 14. We assume that the U.S. meteorologist predicts the same temperature, but expresses it on the Fahrenheit scale (Equation 16.2): T F = + = ° 9 5 15 32 5 ( ) . F 15. I NTERPRET This problem is about converting temperature from the Fahrenheit scale to the Celsius scale. D EVELOP The two temperature scales are related by Equation 16.2: T T F C = + 9 5 32 E VALUATE Solving the above equation for the Celsius temperature, we obtain T T C F = = = ° 5 9 32 5 9 68 32 20 ( ) ( ) C A SSESS This is a useful result to remember, since 20 ° C, or 68 ° F is a typical room temperature. 16. Temperature differences on the Fahrenheit and Celsius scales are related by T T F C = ( / ) , 9 5 so C F . ( )( ) 9 5 10 18 ° = ° (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.) 17. I NTERPRET Given both Fahrenheit and Celsius scales, we want to know when T F and T C are numerically equal. D EVELOP The two temperature scales are related by Equation 16.2: T T F C = + 9 5 32 The condition that the readings are the same numerically is T T T F C C = + = ( ) . 9 5 32 E VALUATE The above equation can be solved to give T T C F = − = − = 5 4 32 40 ( ) A SSESS This is the only temperature in which both scales yield the same reading: ° = − ° 40 40 F C. 18. Equations 16.1 and 16.2 give T C = ° 77 3 273 15 196 . . C, and T F = + = − ° ( )( ) 9 5 196 32 321 F. 19. I NTERPRET This problem is about converting temperature from the Celsius scale to the Fahrenheit scale. D EVELOP The two temperature scales are related by Equation 16.2: T T F C = + 9 5 32. E VALUATE Solving the above equation for the Fahrenheit temperature, we obtain T F = + = ° 9 5 39 1 32 102 4 ( . ) . F A SSESS The temperature is way above the normal body temperature of 98 6 . ° F (or 37 ° C ). Call the doctor immediately! 16

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16.2 Chapter 16 Section 16.2 Heat Capacity and Specific Heat 20. I NTERPRET We find the heat capacity of a large concrete block. We know the mass of the block and its specific heat. D EVELOP The specific heat of concrete is given in Table 16.1 as c = 880 J/kg K. We multiply this by the mass m = 55 000 , kg of the block to find the heat capacity of the block. E VALUATE C mc = = × 48 10 6 J/K. A SSESS This is a large value, but then it takes a large amount of heat to change the temperature of a 55-tonne block of concrete. 21. I NTERPRET We find the energy necessary to change the temperature of an object by a given amount. This involves the heat capacity of the object and the temperature. D EVELOP We use the equation Q mc T = . The mass of the aluminum block is m = 2 0 . kg, the specific heat is c = 900 J/kg K, and the temperature change is T = 18 C°.
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chap 16 - TEMPERATURE AND HEAT 16 TF = EXERCISES Section...

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