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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?

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 60787 Points
3 years ago
Answer ::

a) Use L = L0*(1 + α*ΔT) = 60.0cm*(1 + 1.7x10-6*(85 - 20)) = 60.0063cm

b) Use ΔA = A0*(2*α*ΔT) = π*(0.75cm)2(2*1.7x10-6*(85-20)) = 0.0003903 cm2


Thanks

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