Question

1a) A copper bar is 80cm long at 15°C. What is the increase in length when it is heated to 35°C? The linear expansion coefficient for copper is 1.7×10^-5 °C^-1.

b) A cylinder of diameter 1.00cm at 30°C is to be slid into a hole in a steel plate. The hole has a diameter of 0.99970 cm at 30°C. To what temperature must the plate be heated? The coefficient of linear expansion for steel is 1.1×10^-5 °C^-1.

Answer 1

Answer on Question 80073, Physics, Other

Question:

(a) A copper bar is $80~\mathrm{cm}$ long at $15{}^{\circ}\mathrm{C}$. What is the increase in length when it is heated to $35{}^{\circ}\mathrm{C}$? The linear expansion coefficient for copper is $1.7 \cdot 10^{-5} \, {}^\circ\mathrm{C}^{-1}$.

(b) A cylinder of diameter $1.00~\mathrm{cm}$ at $30{}^{\circ}\mathrm{C}$ is to be slid into a hole in a steel plate. The hole has a diameter of $0.99970~\mathrm{cm}$ at $30{}^{\circ}\mathrm{C}$. To what temperature must the plate be heated? The coefficient of linear expansion for steel is $1.1 \cdot 10^{-5} \, {}^\circ\mathrm{C}^{-1}$.

Solution:

(a) By the definition of the linear thermal expansion we have:

$$\frac{\Delta L}{L_0} = \alpha \Delta T,$$

here, $\Delta L$ is the increase in length of the copper bar when it is heated to $35{}^{\circ}\mathrm{C}$, $L_0$ is the initial length of the copper bar at $15{}^{\circ}\mathrm{C}$, $\alpha$ is the linear expansion coefficient for copper and $\Delta T$ is the change in temperature.

Then, from this equation we can find the increase in length of the copper bar when it is heated to $35{}^{\circ}\mathrm{C}$:

$$\Delta L = \alpha \Delta T L_0 = 1.7 \cdot 10^{-5} \, {}^\circ\mathrm{C}^{-1} \cdot (35{}^{\circ}\mathrm{C} - 15{}^{\circ}\mathrm{C}) \cdot 0.8 \, m = 2.7 \cdot 10^{-4} \, m.$$

(b) By the definition of the linear thermal expansion we have:

$$\frac{\Delta L}{L_0} = \frac{L - L_0}{L_0} = \alpha \Delta T,$$

here, $L_0 = 0.99970~\mathrm{cm}$ is the initial length of the steel plate at $T_0 = 30{}^{\circ}\mathrm{C}$, $L = 1.00~\mathrm{cm}$ is the length of the steel plate at temperature $T$, $\alpha$ is the coefficient of linear expansion for steel and $\Delta T$ is the change in temperature.

Then, we get:

$$\frac{L - L_0}{L_0} = \alpha (T - T_0),$$ $$T = \frac{L - L_0}{\alpha L_0} + T_0 = \frac{1.00 \, \mathrm{cm} - 0.99970 \, \mathrm{cm}}{1.1 \cdot 10^{-5} \, {}^\circ\mathrm{C}^{-1} \cdot 0.99970 \, \mathrm{cm}} + 30{}^{\circ}\mathrm{C} = 57{}^{\circ}\mathrm{C}.$$

Answer:

(a) $\Delta L = 2.7 \cdot 10^{-4} \, m$.

(b) $T = 57{}^{\circ}\mathrm{C}$.

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