Question

To what temperature would you have to heat a brass rod for it to be 2.5 % longer than it is at 30 ∘C?

Express your answer to two significant figures and include the appropriate units.

Answer 1

To make a brass rod 2.5% longer than at 30 °C, one would have to heat it to approximately 1.3×103 °C, or 1346 °C, assuming the coefficient of linear expansion for brass is 19×10-6 °C-1.

Step-by-step explanation:

To find the temperature that would cause a brass rod to expand by 2.5%, we must first calculate the linear expansion using the formula ∆L = L₀α∆T, where ∆L is the change in length, L₀ is the original length, α is the coefficient of linear expansion for brass, and ∆T is the change in temperature (in degrees Celsius). Given that the original length increases by 2.5%, ∆L/L₀ equals 0.025. Re-arranging the formula to solve for ∆T, we get ∆T = (∆L/L₀)/α.

If we assume the coefficient of linear expansion for brass (α) is approximately 19×10-6 °C-1, and knowing the initial temperature is 30 °C, we can solve for the temperature change and add it to the initial 30 °C to get the final temperature required for the 2.5% expansion:

∆T = 0.025 / (19×10-6 °C-1) ≈ 1316 °C

Thus, the rod needs to be heated to:

T = 30 °C + 1316 °C = 1346 °C

To two significant figures, this would be 1.3×103 °C.

Answer 2

By using the coefficient of linear expansion for brass, we can calculate the necessary temperature increase for the rod to be 2.5% longer. This requires heating the brass rod to approximately 1300 \(^{\circ}C\) from an initial temperature of 30 \(^{\circ}C\).

Step-by-step explanation:

First, the linear thermal expansion formula needs to be used to determine the temperature increase required for a brass rod to expand by 2.5%. The formula is:

\(\Delta L = L_{0} \cdot \alpha \cdot \Delta T\)

Where \(\Delta L\) is the change in length, \(L_{0}\) is the original length, \(\alpha\) is the coefficient of linear expansion for brass (approximately \(19 \times 10^{-6} \; ^{\circ}C^{-1}\)), and \(\Delta T\) is the change in temperature in degrees Celsius.

Given that we want the rod to be 2.5% longer, we replace the \(\Delta L\) with \(0.025 \cdot L_{0}\) and solve for \(\Delta T\) as follows:

\(0.025 \cdot L_{0} = L_{0} \cdot \alpha \cdot \Delta T\)

\(\Delta T = \dfrac{0.025}{\alpha}\)

\(\Delta T = \dfrac{0.025}{19 \times 10^{-6}}\)

\(\Delta T \approx 1316 \; ^{\circ}C\)

Since the initial temperature is 30 \(^{\circ}C\), the final temperature (T) would be:

\(T = 30 \; ^{\circ}C + 1316 \; ^{\circ}C\)

\(T \approx 1346 \; ^{\circ}C\)

To two significant figures, the brass rod needs to be heated to approximately 1300 \(^{\circ}C\) to be 2.5% longer than its original length at 30 \(^{\circ}C\).