# ch21 - CHAPTER 21 Thermal Properties and Processes 1* Why...

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CHAPTER 21 Thermal Properties and Processes 1* · Why does the mercury level first decrease slightly when a thermometer is placed in warm water? The glass bulb warms and expands first, before the mercury warms and expands. 2 · A large sheet of metal has a hole cut in the middle of it. When the sheet is heated, the area of the hole will ( a ) not change. ( b ) always increase. ( c ) always decrease. ( d ) increase if the hole is not in the exact center of the sheet. ( e ) decrease only if the hole is in the exact center of the sheet. ( b ) 3 · A steel ruler has a length of 30 cm at 20 ° C. What is its length at 100 ° C? Apply Equ. 21-2. L = (11 × 10 - 6 )(30)(80) cm = 0.0264 cm; L = 30.0264 cm 4 · A bridge 100 m long is built of steel. If it is built as a single, continuous structure, how much will its length change from the coldest winter days (–30 ° C) to the hottest summer days (40 ° C)? Apply Equ. 21-2. L = (11 × 10 - 6 )(100)(70) m = 0.077 m = 7.7 cm 5* ·· ( a ) Define a coefficient of area expansion. ( b ) Calculate it for a square and a circle, and show that it is 2 times the coefficient of linear expansion. ( a ) g = T A/A . ( b ) For a square, A = L 2 (1 + a T ) 2 - L 2 = L 2 (2 a T + a 2 T 2 ) = A (2 a T + a 2 T 2 ); in the limit T 0, A / A = 2 a T , and g = 2 a . For the circle, proceed in same way except that now A = p R 2 ; again, g = 2 a . 6 ·· The density of aluminum is 2.70 × 10 3 kg/m 3 at 0 ° C. What is the density of aluminum at 200 ° C? Apply Equs. 21-4 and 21-5. r = m/V; r = m /( V + V ) = r /(1 + b T ) b = 3 a = 72 × 10 -6 K - 1 r = 2.70 × 10 3 /[1+(72 × 10 - 6 )(200)] = 2.66 × 10 3 kg/m 3

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Chapter 21 Thermal Properties and Processes 7 ·· A copper collar is to fit tightly about a steel shaft whose diameter is 6.0000 cm at 20 ° C. The inside diameter of the copper collar at that temperature is 5.9800 cm. To what temperature must the copper collar be raised so that it will just slip on the steel shaft, assuming the steel shaft remains at 20 ° C? α∆ T = 0.02 cm; use Equ. 21-2 and solve for T T = 0.02/(5.98 × 17 × 10 - 6 ) = 197 C o ; T = 217 o C 8 ·· Repeat Problem 7 when the temperature of both the steel shaft and copper collar are raised simultaneously. Now R Fe = R Cu , and both expand Solve for T and T = (20 + T ) o C 6.0000(1 + 11 × 10 - 6 T ) = 5.9800(1 + 17 × 10 - 6 T ) ) 10 11 (6.00 ) 10 17 (5.98 0.02 6 6 - - × × - × × = T = 561 C o ; T = 581 o C 9* ·· A container is filled to the brim with 1.4 L of mercury at 20 ° C. When the temperature of container and mercury is raised to 60 ° C, 7.5 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the container. 1. Express problem statement in terms of
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## This note was uploaded on 05/01/2008 for the course PHYS 13/14 taught by Professor Millan/thorstensen during the Spring '08 term at Dartmouth.

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ch21 - CHAPTER 21 Thermal Properties and Processes 1* Why...

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