Physics fields and chem combined question

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SUMMARY

The discussion focuses on calculating how long a standard 100W light bulb can be powered using the electrons from a copper penny weighing 5.0g. The current required to operate the bulb is determined to be 0.8333 amps using the formula I = P/V, where P is power and V is voltage. The penny contains approximately 4.68 x 10^21 electrons, resulting in a total charge of 7.488 x 10^2 coulombs. Consequently, the light bulb can run for about 898.6 seconds, or approximately 15 minutes, based on theoretical calculations.

PREREQUISITES
  • Understanding of electrical concepts including current (I), voltage (V), and power (P).
  • Familiarity with atomic mass and molar mass, specifically for copper.
  • Knowledge of basic chemistry, particularly electron charge and Avogadro's number.
  • Ability to apply physics equations such as P = IV and Q = It.
NEXT STEPS
  • Explore the relationship between power, current, and voltage in electrical circuits.
  • Learn about the properties of copper and its applications in electronics.
  • Investigate the concept of charge and its role in electrical systems.
  • Study the practical implications of energy loss in electrical circuits.
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This discussion is beneficial for physics students, electrical engineering students, and anyone interested in the intersection of physics and chemistry in practical applications.

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If a standard 100W light bulb needs 120V to shine, determine how long I could run the light bulb if I used up all the electrons in a copper penny to power it. Assume the penny to have a mass of 5.0g. Hint: think back to the units that make up watts and volts
 
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This is a weird, non-physical question!

I assume they want you to figure out the current flowing through the bulb. Can you do that?

Once you have the current (I), then the total charge (Q) that flows per time (t) is: I = Q/t (that's the meaning of current).

To find the number of electrons in the penny: first find the number of copper atoms. (Hint: you'll need the atomic mass of copper) Then, depending on how wacky your teacher is, find the number of electrons: does he mean ALL electrons? or just the "free" electrons in the outer shell? I would use one electron per atom.

Then consider the charge on the electron. How many "electrons"/sec have to flow to make one Amp of current?
 


To answer this question, we need to use a combination of physics and chemistry concepts. First, let's start with the physics aspect of the question. Watts and volts are both units of power, with watts representing the rate at which energy is used and volts representing the amount of electric potential energy per unit charge. We can use the equation P = IV, where P is power, I is current (measured in amps), and V is voltage, to calculate the amount of power needed to run the light bulb.

In this case, we know that the light bulb needs 100W of power and 120V of voltage. So, using the equation, we can calculate the current needed to power the light bulb as I = P/V = 100W/120V = 0.8333 amps.

Now, let's move on to the chemistry aspect of the question. We are given the mass of a copper penny, which we can use to calculate the number of electrons it contains. Copper has an atomic mass of 63.5 g/mol and a molar mass of 6.022 x 10^23 electrons/mol. So, a 5.0g penny would contain approximately 4.68 x 10^21 electrons.

Next, we need to calculate the total charge of these electrons. Each electron has a charge of 1.6 x 10^-19 coulombs. So, the total charge in the penny would be 4.68 x 10^21 electrons x 1.6 x 10^-19 C/electron = 7.488 x 10^2 C.

Finally, we can use the equation Q = It, where Q is charge, I is current, and t is time, to calculate the time the light bulb could run using the charge in the penny. Plugging in the values we have calculated, we get t = Q/I = 7.488 x 10^2 C/0.8333 A = 898.6 seconds.

Therefore, using all the electrons in a copper penny, we could run the 100W light bulb for approximately 15 minutes (898.6 seconds) before depleting the charge in the penny. It is important to note that this is just a theoretical calculation and in reality, there would be losses in the transfer of energy, so the actual time may be slightly different.
 

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