A 26.0 g copper ring at 0°C has an inner diameter of D = 3.71382 cm. A hollow aluminum sphere at 83.0°C has a diameter of d = 3.72069 cm. The sphere is placed on top of the ring (see the figure), 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 the equilibrium temperature. What is the mass of the sphere? The linear expansion coefficient of aluminum is 23.0 * 10^-6 / C°, the linear expansion coefficient of copper is 17.0 * 10^-6 / C°, the specific heat of aluminum is 900 J/kgK, and the specific heat of copper is 386 J/kgK.

Question

Physics

Scott Campbell

18 November, 20:34

A 26.0 g copper ring at 0°C has an inner diameter of D = 3.71382 cm. A hollow aluminum sphere at 83.0°C has a diameter of d = 3.72069 cm. The sphere is placed on top of the ring (see the figure), 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 the equilibrium temperature. What is the mass of the sphere? The linear expansion coefficient of aluminum is 23.0 * 10^-6 / C°, the linear expansion coefficient of copper is 17.0 * 10^-6 / C°, the specific heat of aluminum is 900 J/kgK, and the specific heat of copper is 386 J/kgK.

+3

Answers (1)

Know the Answer?

Sloane Carey · Answer 1

1.6 g

Explanation:

When they change temperatures their diameters will change following these equations:

D' (t1) = D (t0) * (1 + a (Cu) * (t1 - t0 (Cu)))

d' (t1) = d (t0) * (1 + a (Al) * (t1 - t0 (Al)))

The sphere goes inside the ring when they are at thermal equilibrium (at the same temperature, which is t1). Their diameters will be the same.

D' (t1) = d' (t1)

D (t0) * (1 + a (Cu) * (t1 - t0 (Cu))) = d (t0) * (1 + a (Al) * (t1 - t0 (Al)))

D (t0) + D (t0) * a (Cu) * t1 - D (t0) * a (Cu) * t0 (Cu) = d (t0) + d (t0) * a (Al) * t1 - d (t0) * a (Al) * t0 (Al)

D (t0) * a (Cu) * t1 - d (t0) * a (Al) * t1 = D (t0) * a (Cu) * t0 (Cu) - d (t0) * a (Al) * t0 (Al) - D (t0) + d (t0)

(D (t0) * a (Cu) - d (t0) * a (Al)) * t1 = D (t0) * a (Cu) * t0 (Cu) - d (t0) * a (Al) * t0 (Al) - D (t0) + d (t0)

t1 = (D (t0) * a (Cu) * t0 (Cu) - d (t0) * a (Al) * t0 (Al) - D (t0) + d (t0)) / (D (t0) * a (Cu) - d (t0) * a (Al))

t1 = (3.71382 * 17*10^-6 * 0) - 3.72069 * 23*10^-6 * 83 - 3.71382 + 3.72069) / (3.71382 * 17*10^-6 - 3.72069 * 23*10^-6)

t1 = 10.37 C

To reach this temperature they both exchange heat.

Q (Al) + Q (Cu) = 0

The copper will gain heat (positive) and the aluminum will lose heat (negative)

Q (Al) = - Q (Cu)

mAl * CpAl * (t1 - t0Al) = - mCu * CpCu * (t1 - t0Cu)

mAl = - (mCu * CpCu * (t1 - t0Cu)) / (CpAl * (t1 - t0Al))

mAl = - (0.026 * 386 * (10.37 - 0)) / (900 * (10.37 - 83))

mAl = 1.6*10^-3 kg = 1.6 g