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# 22 - roofner(bar784 Homework 22 Weathers(17104 This...

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roofner (bar784) – Homework 22 – Weathers – (17104) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A square hole 9 . 2 cm along each side is cut in a sheet of copper. Calculate the change in the area of this hole if the temperature of the sheet is increased by 44 K. Correct answer: 0 . 126621 cm 2 . Explanation: Given : L = 9 . 2 cm and Δ T = 44 K . Each side of the hole expands linearly by Δ L = L α Δ T . The area, therefore expands by Δ A = ( L + Δ L ) 2 - A . Since Δ L is much smaller than L we should keep only the linear term in Δ L , so Δ A = 2 L Δ L = 2 α A Δ T = 2 bracketleftbig 1 . 7 × 10 5 ( C) 1 bracketrightbig ( 84 . 64 cm 2 ) (44 K) = 0 . 126621 cm 2 002 (part 1 of 2) 10.0 points At 17 C, an aluminum ring has an inner di- ameter of 6 cm and a brass rod has a diameter of 6 . 01 cm. If the temperature coefficient of expansion for brass is α b = 1 . 9 × 10 5 ( C) 1 and the temperature coefficient of expansion for alu- minum is α a = 2 . 4 × 10 5 ( C) 1 , to what temperature must the ring be heated so that it will just slip over the rod? Correct answer: 86 . 4444 C. Explanation: Given : T 0 = 17 C , d a = 6 cm , d b = 6 . 01 cm , α b = 1 . 9 × 10 5 ( C) 1 , and α a = 2 . 4 × 10 5 ( C) 1 . The new length will be d b = d a (1 + α Δ T ) d b = d a + d a α Δ T T 1 = T 0 + Δ T = T 0 + d b - d a d a α a = 17 C + 6 . 01 cm - 6 cm (6 cm)(2 . 4 × 10 5 ( C) 1 ) = 86 . 4444 C . 003 (part 2 of 2) 10.0 points To what temperature must both be heated so that the ring just slips over the rod? Correct answer: 352 . 458 C. Explanation: We need L Al = L brass for some Δ T . Using the same law, L 0 Al (1 + α a Δ T ) = L 0 brass (1 + α b Δ T ) d a + d a α a Δ T = d b + d b α b Δ T Δ T = T 0 + d b - d a α a d a - α b d b α a d a - α b d b = bracketleftbig 2 . 4 × 10 5 ( C) 1 bracketrightbig (6 cm) - bracketleftbig 1 . 9 × 10 5 ( C) 1 bracketrightbig (6 . 01 cm) = 2 . 981 × 10 5 m / C so Δ T = T 2 - T 0 T 2 = T 0 + Δ T = T 0 + d b - d a α a d a - α b d b = 17 C + 6 . 01 cm - 6 cm 2 . 981 × 10 5 m / C = 352 . 458 C . 004 10.0 points

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roofner (bar784) – Homework 22 – Weathers – (17104) 2 3 mole of xenon gas at 53 C occupies 0 . 154 m 3 . What is the pressure exerted by the Xe atoms on the walls of a container? Correct answer: 52799 . 3 Pa. Explanation: Given : n = 3 mole , T = 53 C = 326 K , V = 0 . 154 m 3 , and R = 8 . 314 J / K · mol .
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