PHysics Question

Question

A certain steel railroad rail is 13 yd in length and weighs 77.4 lb/yd. How much thermal energy is required to increase the length of such a rail by 3.1 mm? Note: Assume the steel has the same specific heat as iron. (Take the average coefficient of linear expansion for steel near room temperature to be 11 ✕ 10⁻⁶ (°C)⁻¹.)

Answer 1

I will work this in algebraic terms and leave the number crunching to you. Start with expansion.

Delta L = change in length = L (alpha) delta T, where L is original length, alpha = expansion coefficient and delta T is the (unknown) temperature change. Solve this guy for delta T.

Now use Q = mC delta T, where Q is thermal energy, m is mass, C is specific heat for iron (you will have to look this up) and delta T is calculated in first step.

Find m = (77.4)(13), plug into the equation and then multiply the string of numbers on the right to get Q the desired answer on the left.

Answer 2

Since the units in this problem are expressed in various systems and the units of SHC for iron is not stated at all, the best approach would be to convert all given quantities to SI units.

Specific heat capacity of iron is 444 Jkg^-1K^-1 = 440Jkg^-1K^-1 (2 sig fig)

13 yards = 13÷1.09=11.89 = 12m (2 sig fig)

total mass (m) of 77.4 lb/yd=1006.2lb ÷ 2.205 = 456.4kg = 456kg (3 sig fig)

ΔL = 3.1mm ÷10³ = 0.0031m

Coefficient of linear expansion of iron is (given) 11x10^-6°C^-1

Now that all units are consistent, we first determine the change in temperature ΔT.

Using the formula ΔL = αLΔT, and solving for ΔT,

ΔT=ΔL/αL

= 0.0031/(11x10^-6x12)=23.48ºC (or K)

Using this value of ΔT in the formula ΔQ=mcΔT, we have

= 456x440x23.48

=4.7x10⁶J (2 sig fig)

Keep in mind that the units were standardized for consistency. You should be able to convert this answer to any other unit for comparison. If you need help with converting or any other problem, let me know.