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A steel ring with a hole having area of 7.970 cm? is to be placed on an aluminum rod with cross-sectional area of 8.000 cm. Both rod and ring are initially at a temperature of 25.0'C. At what common temperature can the steel ring be slipped anto one end of the aluminum rod? (The coefficient of linear expansion of steel is11 10-6 ("c)-1, and the coefficient of linear expansion of aluminum is24 10 ("c)-)

A steel ring with a hole having area of 7.970 cm? is to be placed on an aluminum rod with cross-sectional area of 8.000 cm. Both rod and ring are initially at a temperature of 25.0'C. At what common temperature can the steel ring be slipped anto one end of the aluminum rod? (The coefficient of linear expansion of steel is11 10-6 ("c)-1, and the coefficient of linear expansion of aluminum is24 10 ("c)-)

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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4:20 Access WebAssign EXAMPLE 10.4 Rings and Rods GOAL Apply the equation of area expansion. PROBLEM (a) A circular copper ring at 20.0°C has a hole with an area of 9.980 cm?. What minimum temperature must it have so that it can be slipped onto a steel metal rod having a cross-sectional area of 10.000 em?? (b) Suppose the ring and the rod are heated simultaneously. What minimum change in temperature of both will allow the ring to be slipped onto the end of the rod? (Assume no significant change in the coefficients of linear expansion over this temperature range.) STRATEGY In part (a), finding the necessary temperature change is just a matter of substituting given values into the equation of area expansion. Remember t that y - 20. Part (b) is a little harder because now the rod is also expanding. If the ring is to slip onto the rod, however, the final cross-sectional areas of both ring and rod must be equal, Write this condition in mathematical tems, and solve for AT. SOLUTION (A) Find the temperature of the ring that will allow it to slip onto the rod. Write the equation of area expansion and AA- YAAT 0.020 cm? = [34 x 10 ("cy'9.980 cm?)(AT) substitute known values, leaving AT as the sole unknown Solve for AT, then add this change to the AT = 59°C initial temperature to get the final T= Tg +AT = 20.0°C + 59°C = 79'C temperature. (B) If both ring and rod are heated, find the minimum change in temperature that will allow the ring to be slipped onto the rod. Set the final areas of the copper ring and Ac + AAc = Ag + AAS steel rod equal to each other. خي ي + A ي الم +A A وAATي = - آش هY (YAc - YAg) AT Ag - Ac Substitute for each change in area, AA. Rearrange terms to get AT on one side only, factor it out, and solve. As - Ac AT = 10.000 cm2 - 9.980 cm? (34 x 10-6 "c(9.980 cm) - (22 x 10-6-c10.000 cm) AT - 170°C LEARN MORE REMARKS Warming and cooling strategies are sometimes useful for separating glass parts in a chemistry lab, such as the glass stopper in a bottle of reagent. QUESTION If instead of heating the copper ring in part (a) the steel rod is cooled, would the magnitude of the required temperature change be larger, smaller, or the same? Why? (Dont calculate it') O smaller, because steel has a larger coefficient of area expansion O larger, because the steel rod has a smaler radius O larger, because steel has a smaller coefficient of area expansion Osmaller, because the steel rod has a smaller radius PRACTICE IT Use the worked example above to help you solve this problem. (a) A circular copper ring at 22.0°C has a hole with an area of 9.95 cm. What minimum temperature must it have so that it can be slipped onto a steel metal rod having a cross-sectional area of 10.0 cm?? 5. "C (b) Suppose the ring and the rod are heated simultaneously. What change in temperature of both will allow the ring to be slipped onto the end of the rod? (Assume no significant change in the coefficients of linear expansion over this temperature range.) "C EXERCISE HINTS: GETTING STARTED I IM STUCK! A steel ring with a hole having area of 7.970 cm2 is to be placed on an aluminum rod with cross-sectional area of 8.000 cm. Both rod and ring are initially at a temperature of 25.0°C. At what common temperature can the steel ring be slipped onto one end of the aluminum rod? (The coefficient of linear expansion of steel is11 x 10-6 ("c)-1, and the coefficient of linear expansion aof aluminum is24 - 10 ("C)-1) "C II
Transcribed Image Text:4:20 Access WebAssign EXAMPLE 10.4 Rings and Rods GOAL Apply the equation of area expansion. PROBLEM (a) A circular copper ring at 20.0°C has a hole with an area of 9.980 cm?. What minimum temperature must it have so that it can be slipped onto a steel metal rod having a cross-sectional area of 10.000 em?? (b) Suppose the ring and the rod are heated simultaneously. What minimum change in temperature of both will allow the ring to be slipped onto the end of the rod? (Assume no significant change in the coefficients of linear expansion over this temperature range.) STRATEGY In part (a), finding the necessary temperature change is just a matter of substituting given values into the equation of area expansion. Remember t that y - 20. Part (b) is a little harder because now the rod is also expanding. If the ring is to slip onto the rod, however, the final cross-sectional areas of both ring and rod must be equal, Write this condition in mathematical tems, and solve for AT. SOLUTION (A) Find the temperature of the ring that will allow it to slip onto the rod. Write the equation of area expansion and AA- YAAT 0.020 cm? = [34 x 10 ("cy'9.980 cm?)(AT) substitute known values, leaving AT as the sole unknown Solve for AT, then add this change to the AT = 59°C initial temperature to get the final T= Tg +AT = 20.0°C + 59°C = 79'C temperature. (B) If both ring and rod are heated, find the minimum change in temperature that will allow the ring to be slipped onto the rod. Set the final areas of the copper ring and Ac + AAc = Ag + AAS steel rod equal to each other. خي ي + A ي الم +A A وAATي = - آش هY (YAc - YAg) AT Ag - Ac Substitute for each change in area, AA. Rearrange terms to get AT on one side only, factor it out, and solve. As - Ac AT = 10.000 cm2 - 9.980 cm? (34 x 10-6 "c(9.980 cm) - (22 x 10-6-c10.000 cm) AT - 170°C LEARN MORE REMARKS Warming and cooling strategies are sometimes useful for separating glass parts in a chemistry lab, such as the glass stopper in a bottle of reagent. QUESTION If instead of heating the copper ring in part (a) the steel rod is cooled, would the magnitude of the required temperature change be larger, smaller, or the same? Why? (Dont calculate it') O smaller, because steel has a larger coefficient of area expansion O larger, because the steel rod has a smaler radius O larger, because steel has a smaller coefficient of area expansion Osmaller, because the steel rod has a smaller radius PRACTICE IT Use the worked example above to help you solve this problem. (a) A circular copper ring at 22.0°C has a hole with an area of 9.95 cm. What minimum temperature must it have so that it can be slipped onto a steel metal rod having a cross-sectional area of 10.0 cm?? 5. "C (b) Suppose the ring and the rod are heated simultaneously. What change in temperature of both will allow the ring to be slipped onto the end of the rod? (Assume no significant change in the coefficients of linear expansion over this temperature range.) "C EXERCISE HINTS: GETTING STARTED I IM STUCK! A steel ring with a hole having area of 7.970 cm2 is to be placed on an aluminum rod with cross-sectional area of 8.000 cm. Both rod and ring are initially at a temperature of 25.0°C. At what common temperature can the steel ring be slipped onto one end of the aluminum rod? (The coefficient of linear expansion of steel is11 x 10-6 ("c)-1, and the coefficient of linear expansion aof aluminum is24 - 10 ("C)-1) "C II
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