Question

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?

Answer #1

**at equilibrium temperature**

**diameter of sphere = diameter of ring**

**d*(1 + alpha_A*(100-t)) = D*(1+ alph_C*(t-0))**

**alpha_A = coefficeint of expansion of Aluminium =
22.2*10^-6**

**alpha_C = coefficeint of expansion of copper =
16.6*10^-6**

**2.46507*(1-(22.2*10^-6*(100-t))) =
2.46*(1+(16.6*10^-6*t))**

**equilibrium temperature t = 28.98 oC**

**=============**

**heat lost by aluminium = heat gained by copper**

**Mal*Sal*(100-t) = Mc*Sc*(0-t)**

**specific heat of aluminium = Sal = 910 J/kg
K**

**specific heat of copper = Sc = 386 J/kg K**

**Mal*910*(100-28.98) = 0.023*386*28.98**

**Mal = 0.00398 kg = 3.98 grams**

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?

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
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steel and aluminum are 12×10−6 K−1 and 23×10−6 K−1
respectively.

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
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To what common temperature should the ring and the shaft be
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23×10-6K-1 respectively.
Express your answer in degrees Celsius to two significant
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show all work

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