An 8 000-kg aluminum flagpole 100-m high is heated by the sun from a temperature of 10$\degree$C to 20$\degree$C.Find the increase in internal energy (in J)of the aluminum.(The coefficient of linear expansion is 24 * 10 ^- ⁶ ($\degree$C) ^- ¹,the density is 2.7* 10³ kg/m³,and the specific heat of aluminum is 0.215 cal/g.$\degree$C. )

Question

Quizplus · Accepted Answer

We can use the equation for the change in internal energy of a substance:

ΔU = mcΔT

where ΔU is the change in internal energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, we need to find the change in temperature in kelvin:

ΔT = 20°C - 10°C = 10 K

Next, we can find the mass of the flagpole:

m = ρV = ρAh

where ρ is the density, A is the cross-sectional area, and h is the height.

A circle with a diameter of 1 meter has an area of π/4 m^2. So the cross-sectional area is:

A = (π/4)(1.5 m)^2 = 1.77 m^2

Now, we can find the volume:

V = Ah = (1.77 m^2)(100 m) = 177 m^3

And finally, the mass:

m = (2.7×10³ kg/m³)(177 m³) = 478.3×10³ kg

Now we can calculate the change in internal energy:

ΔU = mcΔT = (478.3×10³ kg)(0.215 cal/g°C)(10 K)(4.184 J/cal)

ΔU = 7.22×10¹⁷ J ≈ 7.2×10¹⁰ J

So the best choice is B.

An 8 000-Kg Aluminum Flagpole 100-M High Is Heated by the Sun