16_InstSolManual_PC

# 16_InstSolManual_PC - TEMPERATURE AND HEAT 16 9 TF = TC 32 5 EXERCISES Section 16.1 Heat Temperature and Thermodynamic Equilibrium 14 15 We assume

This preview shows pages 1–3. Sign up to view the full content.

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): 9 5 ( 15) 32 5 F. F T = - + = ° 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: 9 32 5 F C T T = + E VALUATE Solving the above equation for the Celsius temperature, we obtain 5 5 ( 32) (68 32) 20 C 9 9 C F T T = - = - = ° 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 (9/5) , F C T T = 9 5 so ( )(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.) 17. I NTERPRET Given both Fahrenheit and Celsius scales, we want to know when F T and C T are numerically equal. D EVELOP The two temperature scales are related by Equation 16.2: 9 32 5 F C T T = + The condition that the readings are the same numerically is 9 5 ( ) 32 . F C C T T T = + = E VALUATE The above equation can be solved to give 5 (32) 40 4 C F T T = - = - = A SSESS This is the only temperature in which both scales yield the same reading: 40 F 40 C. - ° = - ° 18. Equations 16.1 and 16.2 give 77.3 273.15 196 C, C T = - - ° . and 9 5 ( )( 196) 32 321 F. F T = - + = - ° 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: 9 5 32. F C T T = + E VALUATE Solving the above equation for the Fahrenheit temperature, we obtain 9 (39.1) 32 102.4 F 5 F T = + = ° A SSESS The temperature is way above the normal body temperature of 98.6 F ° (or 37 C ° ). Call the doctor immediately! 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. 16.1 16

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
16.2 Chapter 16 D EVELOP The specific heat of concrete is given in Table 16.1 as 880 J/kg K. c = We multiply this by the mass 55,000 kg m = of the block to find the heat capacity of the block. E VALUATE 6 48 10 J/K. C mc = = × 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
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.

### Page1 / 15

16_InstSolManual_PC - TEMPERATURE AND HEAT 16 9 TF = TC 32 5 EXERCISES Section 16.1 Heat Temperature and Thermodynamic Equilibrium 14 15 We assume

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online