Question

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C ? Coefficient of linear expansion of copper = 1.70 x 10-5K-1.

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Answer 1

Answer:

0.0144 cm is the change in the diameter of the hole(increases).

Explanation:

∆A/A={(d2)^2-(d1)^2}/d1^2=beta(T2-T1)=2alpha(T2-T1)

when you substitute you get d2=4.2544

d2-d1=4.2544-4.24=0.0144cm.

Answer 2

Answer:

Initial temperature, T1 =27.0 °c

Diameter of the hole at T1 ,d1 =4.24cm

Final temperature, T2=227 °c

Diameter of the hole at T2=D2

Co-efficient of linear expansion of copper,

αcu=1.70×10 −5 K −1

For co-efficient of superficial expansion β,and change in temperature △T, we have the relation:

Change in area (△) / Original area (A) = β△T

[(πd 22/4)−(πd 22/4)]/(πd 11 /4)=△A/A

∴△A/A=(d 22 −d 12 )/d 12

But β=2α

∴(d22 −d 12 )/d 12=2α△T

(d 22 /d 12 )−1=2α(T2−T1)

d22/4.24 2 =2×1.7×10−5 (227−27)+1

d22 =17.98×1.0068=18.1

∴d2 =4.2544cm

Change in diameter =

d2 − F1 =4.2544−4.24=0.0144cm

Hence, the diameter increases by 1.44×10 −2cm

Explanation:

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