Would a grounded copper heat sink located near a room's ceiling be an effective way to reduce a room's temperature?

I am thinking about a cheap but effective way to reduce the temperature of my bedroom in my apartment during hot summer nights. I live in an old apartment building and the air conditioner located in the living room is old and it can make a lot of noise at times, which makes it very hard to fall asleep. This often forces me turn it off and to turn on a fan, yet hot air blowing around the bedroom doesn't do much to cool me down and I still have trouble falling asleep.

What I am thinking about doing is using a grounded copper heat sink to channel heat out of my bedroom. I plan to mount a 8 foot long, 1/2" diameter copper pipe to the bedroom's ceiling horizontally so that the pipe lies flush against the ceiling. Then I plan to attach one end of a copper wire to the pipe and then attach the other end of the copper wire to the end of a special grounding cord, one which uses a standard 3-prong AC connector (reference picture below) to plug into an AC outlet.

(Note: I want to point out that a standard AC power cord which has three wires cannot be used for this experiment due to the risk of electrical shock. Only a special grounding cord like this one can be used.)

So, since copper is an excellent conductor of heat, I am thinking that a lot of the heat within the hot air sitting along the ceiling of my bedroom will be absorbed into the copper pipe and then that heat will be channeled down the copper wire, then into the grounding cord, then into the ground wire of the AC wiring within the walls, and then eventually down into the grounding rod within the ground.

Would a grounded copper heat sink located near a room's ceiling be an effective way to reduce a room's temperature?

• Definitely a candidate for what-if.xkcd. Randall would take this and run in scary directions. May 13, 2022 at 15:11

We need Fourier's law of heat conduction $$q = - k \nabla T\,,$$ which in a wire of length $$L$$ and cross-section $$A$$ with an assumed homogeneous gradient gives the total heat flux out of your room $$Q = q A = \frac{A k \Delta T}{L}\,,$$ with the temperature difference between the ends of the wire (ideally your room temperature and ground temperature or something like that).
Let's give you room $$30\,$$m$$^3$$ of air and calculate the time to lower the temperature by $$1\,$$K. This requires $$3\cdot10^4\,$$J to escape (calculated with the heat capacity of air), and estimating a wire of 10$$\,$$m length and 10$$\,$$mm diameter plus a temperature difference of 10$$\,$$K between inside and outside the total heat flux through the wire would be so small that it would take on the order of 10 days to cool your room by 1$$\,$$K.
• In the calculation above this lowers $L$ by about a factor of 10 and we are down to 1 day... For a concrete setup you can do the maths yourself. May 13, 2022 at 14:11