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

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16.1 TEMPERATURE AND HEAT EXERCISES 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. INTERPRET This problem is about converting temperature from the Fahrenheit scale to the Celsius scale. DEVELOP The two temperature scales are related by Equation 16.2: TT FC =+ 9 5 32 EVALUATE Solving the above equation for the Celsius temperature, we obtain CF =− = ° 5 9 32 5 9 68 32 20 C ASSESS 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 ∆∆ = (/) , 95 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. INTERPRET Given both Fahrenheit and Celsius scales, we want to know when T F and T C are numerically equal. DEVELOP The two temperature scales are related by Equation 16.2: 9 5 32 The condition that the readings are the same numerically is T C = . 9 5 32 EVALUATE The above equation can be solved to give = 5 4 32 40 ASSESS This is the only temperature in which both scales yield the same reading: −°= −° 40 40 . 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. INTERPRET This problem is about converting temperature from the Celsius scale to the Fahrenheit scale. DEVELOP The two temperature scales are related by Equation 16.2: 9 5 32. EVALUATE Solving the above equation for the Fahrenheit temperature, we obtain T F = ° 9 5 39 1 32 102 4 (. ) .F ASSESS 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. INTERPRET We find the heat capacity of a large concrete block. We know the mass of the block and its specific heat. DEVELOP 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 ,k g of the block to find the heat capacity of the block. EVALUATE Cm c == × 48 10 6 J/K. ASSESS 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. INTERPRET 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. DEVELOP We use the equation Qm cT =∆ . The mass of the aluminum block is m = 20 . kg, the specific heat is c 900 J/kg K, and the temperature change is T = 18 C°. EVALUATE 32 000 ,J . ASSESS The same value would be the heat released by the aluminum if it cooled 18 C°.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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

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