Question

At 20∘C, the hole in an aluminum ring is 2.200 cm in diameter. You need to slip this ring over a steel shaft that has a room-temperature diameter of 2.204 cm .

To what common temperature should the ring and the shaft be heated so that the ring will just fit onto the shaft? Coefficients of linear thermal expansion of steel and aluminum are

12×10^{-6}K^{-1} and
23×10^{-6}K^{-1} respectively.

Express your answer in degrees Celsius to two significant figures.

Answer #1

The linear thermal expansion is given by the following equation,

ΔL = αLo ΔT

where ΔL is the change in length L, ΔT is the change in
temperature, and α is the coefficient of **linear
expansion**,

the linear expansion coefficient of steel is, α = 12 x 10^-6 K^-1 and that of the aluminum is α = 23 x 10^-6 K^-1.

L = Lo(1 + αΔT)

The initial temperature is 20°C. If the final common temperature
is T°C, then ΔT = T-20°

At temperature T, the diameter of the hole in the aluminium
is:

2.200(1 + 23×10^−6(T-20))

At temperature T, the diameter of the steel shaft is:

2.204(1 + 12×10^−6(T-20))

These must be equal so:

2.200[1 + 23×10^−6(T-20)] = 2.204[1 + 12×10^−6(T-20)]

The common temperature at which the ring and the shaft be heated so that the ring will just fit onto the shaft is,

T = 35650°C = 3600°C = **3.6** ×**10^3° C
_Ans.(2 significant figures)**

At 20∘C, the hole in an aluminum ring is 2.100 cm in diameter.
You need to slip this ring over a steel shaft that has a
room-temperature diameter of 2.107 cm To what common temperature
should the ring and the shaft be heated so that the ring will just
fit onto the shaft? Coefficients of linear thermal expansion of
steel and aluminum are 12×10−6 K−1 and 23×10−6 K−1
respectively.

A brass ring with inner diameter 2.00 cm and outer diameter 3.00
cm needs to fit over a 2.00-cm-diameter steel rod, but at 20
degrees Celsius the hole through the brass ring is 40 μm too small
in diameter.
To what temperature, in degrees Celsius, must the rod and ring
be heated so that the ring just barely slips over the rod?

A circular aluminum [a = 23.0 × 10−6 /°C]
ring is to be mounted on a circular steel tube. The ring will be
heated so that it expands enough to slip onto the steel tube. After
the ring cools, there will be a tight fit between the aluminum ring
and the steel tube. (This process is called shrink
fit.)
The outside diameter of the steel tube is 440 mm, and the inside
diameter of the aluminum ring is 438 mm....

2 parts thx!
1. A flat aluminum plate has a circular hole of diameter 1.739
cm at 0.00 °C.What is the change in its diameter when the
temperature of the plate is increased to 116 °C? For reference, the
coefficient of thermal expansion for aluminum is 2.3 x
10-5/C°. Give at least 2 significant figures in your
answer, which should be in cm.
2. A flat plate made of an unknown metal has a circular hole of
diameter 2.89 cm...

A brass ring of diameter 10.00 cm at 22.0°C is heated and
slipped over an aluminum rod of diameter 10.01 cm at 22.0°C. Assume
the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to
separate the two metals?
°C
(b) What if the aluminum rod were 10.02 cm in diameter?
°C

A brass ring of diameter 10.00 cm at 15.7°C is heated and
slipped over an aluminum rod of diameter 10.01 cm at 15.7°C. Assume
the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to
separate the two metals?
°C
Is that temperature attainable?
Yes No
(b) What if the aluminum rod were 10.04 cm in diameter?
°C
Is that temperature attainable?

A brass ring of diameter 10.00 cm at 17.8°C is heated and
slipped over an aluminum rod of diameter 10.01 cm at 17.8°C. Assume
the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to
separate the two metals?
°C
(b) Is that temperature attainable?
Yes or No
(c) What if the aluminum rod were 9.16 cm in diameter?
°C
(d) Is that temperature attainable?
Yes or No

A steel rod is 3.000 cm in diameter at 33.0 °C. A brass ring has
an interior diameter of 2.992 cm at 33.0 °C. At what common
temperature will the ring just slide onto the rod? The coefficients
of linear expansion of steel and brass are 1.100×10-5 , 1.900×10-5
/°C, respectively, answer in °C.

A steel rod is 2.689 cm in diameter at 35.00°C. A brass ring has
an interior diameter of 2.684 cm at 35.00°C. At what common
temperature will the ring just slide onto the rod? The linear
expansion coefficient of steel is 11.00 × 10-6 1/C°. The
linear expansion coefficient of brass is 19.00 × 10-6
1/C°.
4sig figs

At 18.10 ∘C a brass sleeve has an inside diameter of 2.22063 cm
and a steel shaft has a diameter of 2.22333 cm. It is desired to
shrink-fit the sleeve over the steel shaft.
A.) To what temperature must the sleeve be heated in order for
it to slip over the shaft?
B.) Alternatively, to what temperature must the shaft be cooled
before it is able to slip through the sleeve?

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 15 minutes ago

asked 17 minutes ago

asked 24 minutes ago

asked 24 minutes ago

asked 34 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago