Q:

Help me solve this TAOE problem

So, I picked up this book and am working through it by myself. I am a bit stuck on problem 1.6:

In the bottom you can see how I'm approaching this. I'm assuming the transmission cable goes across NY, and New Yorkers draw power from different points along the line. Since they each draw power at 115V, then the total voltage at the station, V@Station minimum, is 115 x 107V.
The power dissipated through one foot of cable is 
(V@Station ^ 2) / Resistance per foot of cable

This worked out to be 2.645 x 1025W. And 1010W would be dissipated in 3.78 x 10-16 feet.
I haven't calculated the temperature because I don't know how to. 
The last part of the question comments on these strange  results and asks for a resolution to this problem. I'm guessing that, since power rises non linearly with voltage, we reduce the V@Station by the use of transformers in order to reduce power dissipation through the transmission cable. 
Is this approach and the answer correct? Please add your own thoughts and point me in the right direction if I'm wrong.

A:

(a)
If we want to supply \$10^{10}\$ watts of power at \$115\$ Volts then the current would have to be 
$$ I=\frac{10^{10}}{115} = 86.956 \times 10^6 Amperes$$ .
Power lost per foot = \$I^2\$ (Resistance per foot)
$$\frac{P}{foot} = I^2\frac{R}{foot} = 7.561\times 10^{!5} \times 0.05 \times 10^{-6} $$
$$= 378.071 \times 10^6 \text{ Watt per foot}$$
(b)
So all the power \$10^{10}\$ is lost in 
$$\frac{10^{10}}{78.071 \times 10^6} = 26.45 \text{ feet} \approx 8 \text{ metres}$$
(c)
The value of the Stefan–Boltzmann constant is given in SI units by
$$σ = 5.670367×10^{−8} \frac{W}{m^{2} K^{4}}$$
Emissivity of copper \$ε = 0.023\$. Note that the diameter of the wire is \$0.3048\$ metres.
Surface area of cylindrical wire 
$$A = 2\pi rh = 0.3048 \times \pi \times 8.06196 = 7.71 m^2$$
Power radiated from heated body 
$$= ε σ A T^4 = 1.0068 \times 10^{-8}T^4 = I^2R = 10^{10}$$
$$T^4 = 9.932 \times 10^{17}$$
$$T=31,569.22 \text{ Kelvin}$$
To put that into perspective, copper melts at \$1,358\$ Kelvin while the surface of the sun is \$5,778\$ Kelvin hot.

A:

As hinted by others in comments, that question is formulated so that you might reflect on the benefits of high-voltage power transmission lines. 
Power is not transmitted over long distances at such a low voltage as 115Vac (230-240Vac in EU). This latter, relatively low, voltage is used in households for safety and practical reasons, but it is wasteful for long range transmission. You may find some mathematical result in this section of the article linked above.

