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A piece of copper tubing (αcoper = 17 × 10−6/ ◦C) used in plumbing has a length of 60.0 cm and an inner diameter of 1.50 cm at 20◦ C. When hot water at 85◦ C flows through the tube, what are (a) the tube’s new length and (b) the change in its cross sectional area? A piece of copper tubing (αcoper = 17 × 10−6/ ◦C) used in plumbing has a length of 60.0 cm and an inner diameter of 1.50 cm at 20◦ C. When hot water at 85◦ C flows through the tube, what are (a) the tube’s new length and (b) the change in its cross sectional area?
Answer ::a) Use L = L0*(1 + α*ΔT) = 60.0cm*(1 + 1.7x10-6*(85 - 20)) = 60.0063cmb) Use ΔA = A0*(2*α*ΔT) = π*(0.75cm)2(2*1.7x10-6*(85-20)) = 0.0003903 cm2Thanks
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