Chapter 16, Hm. 10 Solutions

Chapter 16, Hm. 10 Solutions - Solutions to Homework #10...

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Solutions to Homework #10 Chapter 16 Picture the Problem : A temperature difference is given in Fahrenheit degrees and needs to be converted to Celsius degrees and to kelvins. Strategy: Write the temperature difference in Fahrenheit as a final temperature minus the initial temperature. Use equation 16-1 to write an equation to relate the temperature difference in Fahrenheit with the corresponding temperature difference in Celsius. Do the same with equation 16-3 to find the conversion between Celsius and Kelvin. Solution: 1. (a) Find the temperature difference in Celsius: ( 29 ( 29 ( 29 ( 29 5 9 C C2 C1 2 F1 9 5 5 5 F2 F1 F 9 9 5 9 32 °F 32 °F 24 F 13 C° F T T T T T T T T = - = - - - = - = ∆ = ° = 2. (b) Find the temperature difference in Kelvin: ( 29 ( 29 K 2 1 2 1 2 1 C 273.15 273.15 13 K K K C C C C T T T T T T T T = - = + - + = - = ∆ = Insight: Since the Celsius degree and the Kelvin degree have the same size, a change in Celsius has the same magnitude as the same change in Kelvin. Picture the Problem A brass sleeve has an inner diameter slightly smaller than the diameter of a steel bar. To shrink-fit the sleeve over the bar, you must either heat the sleeve or cool the bar. Strategy: Solve equation 16-4 for the temperature at which the change in diameter is equal to the difference in diameters of the brass sleeve and the steel rod. For the case of heating the brass sleeve use the coefficient of thermal expansion of brass, the initial inner diameter of the brass and a positive change in diameter. For the case of cooling the steel rod, use the coefficient of thermal expansion of steel, the diameter of the steel rod, and a negative change in diameter. The coefficients of thermal expansion of brass and steel aluminum are given in Table 16-1. Solution: 1 . Solve equation 16-4 for the final temperature: ( 29 0 0 L L T L T T L T T L α ∆ = = - = + 2. (a) Insert the data for heating the brass sleeve: ( 29 ( 29 5 1 2.19893 cm 2.19625 cm 12.25 C 76 C 1.9 10 K 2.19625 cm T - - - = ° + = ° ×
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3. (b) Insert the data for cooling the steel: ( 29 ( 29 –6 1 2.19625 cm – 2.19893 cm 12.25 C 89 C 12 10 K 2.19893 cm T - = ° + = - ° × Insight: Since the coefficient of thermal expansion for brass is greater than the coefficient of thermal expansion for steel, the brass does not have to be heated through as large of a temperature difference as the steel has to be cooled to achieve the same change in diameter. Picture the Problem : A person is lifting weights during a workout. The person does work against gravity each time the weight is lifted. Strategy: Calculate the amount of work done each time the weight is lifted and convert the results to calories. Divide the total work done by the work per lift to calculate the number of lifts necessary to expend the specified amount of calories. Solution:
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This note was uploaded on 10/17/2011 for the course PHY 101 taught by Professor Ashkenkai during the Fall '08 term at FIU.

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Chapter 16, Hm. 10 Solutions - Solutions to Homework #10...

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