Question

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?

Answer #1

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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
12×10-6K-1 and
23×10-6K-1 respectively.
Express your answer in degrees Celsius to two significant
figures.

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 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 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 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 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

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?

A 23.0 g copper ring at 0.000°C has an inner diameter of
D = 2.46000 cm. An aluminum sphere at 100.0°C has a
diameter of d = 2.46507 cm. The sphere is put on top of
the ring, and the two are allowed to come to thermal equilibrium,
with no heat lost to the surroundings. The sphere just passes
through the ring at equilibrium temperature.
What is the mass of the sphere?

A 23.0 g copper ring at 0.000°C has an inner diameter of
D = 2.46000 cm. An aluminum sphere at 100.0°C has a
diameter of d = 2.46507 cm. The sphere is put on top of
the ring, and the two are allowed to come to thermal equilibrium,
with no heat lost to the surroundings. The sphere just passes
through the ring at equilibrium temperature.
What is the mass of the sphere?

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