What is the change in temperature of a coin dropped from a building?

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The discussion centers on calculating the change in temperature of a 5-gram copper coin dropped from a 300-meter building, reaching a terminal velocity of 45 m/s. The conservation of energy principle is applied, where the initial potential energy is equated to the sum of the heat energy (delta Q) and kinetic energy (KE). The final calculation yields a temperature change of approximately 4.98 °C, confirming that the units for temperature are correctly derived as °C after canceling out kg and J.

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Homework Statement



A 5-g coin is dropped from a 300-m building. If it reaches a terminal velocity of 45 m/s, and the rest of the energy is converted to heating the coin, what is the change in temperature (in °C) of the coin? (The specific heat of copper is 387 J/kg °C.)

I have no idea how to start the problem..
 
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Think about conservation of energy.
 
So...
Potential Energy initial = delta Q?

I don't have a physics book and the teacher pdf notes kinda suck can i get the formula for the thermal change in energy
 
intenzxboi said:
So...
Potential Energy initial = delta Q?

Nope. Not all of it is converted into heat. Obviously if the object starts at 0 m/s and ends up at 45 m/s, it has gained some kinetic energy too.
 
mgd= delta q + KE?
 
No
Total energy before the coin is dropped = total energy after it reaches terminal velocity
What is/are the energy/energies present before it is dropped?
What is/are the energy/energies present after it reaches terminal velocity?
 
Sorry I'm late...
the last one is right
 
intenzxboi said:
mgd= delta q + KE?

Yes.
 
(.005)(9.8)(300) = (5)(387)T + (.5)(.005)(45^2)

So t = 4.98
does that look right?
 
  • #10
my other question is if i did it like that wouldn't the units for my answer be 4.98 kg/J = delta T ??
 
  • #11
the number is right but not the unit. What is the unit for temperature?
 
  • #12
o0o got it C the kg and J cancels out
 
  • #13
thanks for all the help
 

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