Question

The brass bar and the aluminum bar in the drawing are each attached to an immovable wall. At 21.9 °C the air gap between the rods is 1.55 x 10-3 m. At what temperature will the gap be closed?

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Answer 1

Answer on Question#37950 - Physics - Other

The brass bar and the aluminum bar in the drawing are each attached to an immovable wall. At $21.9{}^{\circ}\mathrm{C}$ the air gap between the rods is $1.55 \times 10^{-3} \mathrm{~m}$. At what temperature will the gap be closed?

Solution:

$\Delta \mathrm{L} = \alpha \mathrm{L}_0\Delta \mathrm{T}$ gives for the expansion of the aluminum

$$\Delta \mathrm {L} _ {\mathrm {A}} = \alpha_ {\mathrm {A}} \mathrm {L} _ {\mathrm {A}} \Delta \mathrm {T}$$

and the expansion of the brass

$$\Delta \mathrm {L} _ {\mathrm {B}} = \alpha_ {\mathrm {B}} \mathrm {L} _ {\mathrm {B}} \Delta \mathrm {T}$$

Taking the coefficients of thermal expansion for aluminum $(\alpha_{\mathrm{A}} = 23\times 10^{-6}\mathrm{K}^{-1})$ and brass $(\alpha_{\mathrm{B}} = 19\times 10^{-6}\mathrm{K}^{-1})$ adding Equations (1) and (2), and solving for $\Delta T$ give:

$$\Delta \mathrm {T} = \frac {\Delta \mathrm {L} _ {\mathrm {A}} + \Delta \mathrm {L} _ {\mathrm {B}}}{\alpha_ {\mathrm {A}} \mathrm {L} _ {\mathrm {A}} + \alpha_ {\mathrm {B}} \mathrm {L} _ {\mathrm {B}}} = \frac {1 . 5 5 \times 1 0 ^ {- 3} \mathrm {m}}{2 3 \times 1 0 ^ {- 6} \mathrm {K} ^ {- 1} \cdot 1 \mathrm {m} + 1 9 \times 1 0 ^ {- 6} \mathrm {K} ^ {- 1} \cdot 2 \mathrm {m}} = 2 5. 4 {}^ {\circ} \mathrm {C}$$

The desired temperature is then

$$\mathrm {T} = 2 1. 9 {}^ {\circ} \mathrm {C} + 2 5. 4 {}^ {\circ} \mathrm {C} = 4 7. 3 {}^ {\circ} \mathrm {C}$$

Answer: the gap will be closed at temperature $47.3{}^{\circ}\mathrm{C}$.