A brass plug is to be placed in a ring made of iron. At 15 C, the diameter of th

Question

A brass plug is to be placed in a ring made of iron. At 15 C, the diameter of the plug is 8.712 cm and that of the inside of the ring is 8.700 cm .

Part A

They must both be brought to what common temperature in order to fit?

Express your answer using two significant figures.

Part B

What if the plug were iron and the ring brass?

Express your answer using two significant figures.

A brass plug is to be placed in a ring made of iron. At 15 C, the diameter of the plug is 8.712 cm and that of the inside of the ring is 8.700 cm .

Part A

They must both be brought to what common temperature in order to fit?

Express your answer using two significant figures.

Part B

What if the plug were iron and the ring brass?

Express your answer using two significant figures.

Explanation / Answer

Assuming constant coefficient of thermal expansion and small thermal strain, diameter of the plug and the ring can be found from the formula:
d = d + d = d(1 + T)

Since both objects undergo the same change of temperature from 15°C to another common temperature, you can write:
di = di(1 + iT) (for iron ring)
db = db(1 + bT) (for brass plug)

At the temperature we're are looking both diameters are the same,i.e.:
di = db

di(1 + iT) = db(1 + bT)

di + diiT = db + dbbT
Solve for T
T = (db - di) / (dii - dbb)
= (8.712cm - 8.70cm) / (8.70cm12×10K¹ - 8.712cm19×10K¹)
= -196K = -196°C

So the common temperature they need to be brought in order to fit is:
T = T + T = 15°C - 196°C = -180°C

(b)T = (db - di) / (dii - dbb)

(8.7cm - 8.712cm) / ( 8.712cm12×10K1 -8.70cm19×10K¹ ) =198 o C

So the common temperature they need to be brought in order to fit is:
T = T + T = 15°C + 198 oC =213 = 210 oC

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