- #1
noboost4you
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"A piece of aluminum (specific heat 910 J/kg-C) of mass 200g at 80.0 deg Celcius is dropped into a styrofoam cup filled with 100mL of water at 20.0 deg C. What are the final temperatures of the water and the aluminum?"
This is what I got:
I figured out the amount of heat energy transferred for each using the equation Q = mc(delta T)
a)
m = 200g = .200kg
c = 910 J/kg-C)
T = 80.0 deg C
Q = (.200kg)(910 J/kg-C)(80.0C)
Q = 14560 J
b)
m = 100mL = 100g = .100kg
c = 4190 J/kg-C)
T = 20.0 deg C
Q = (.100kg)(4190 J/kg-C)(20.0C)
Q = 8380 J
Ok, now this is where I'm confused. I don't know how to incorporate both of the temperatures when they are interacting with each other, or even what I did above was necessary.
Any suggestions? Thanks
This is what I got:
I figured out the amount of heat energy transferred for each using the equation Q = mc(delta T)
a)
m = 200g = .200kg
c = 910 J/kg-C)
T = 80.0 deg C
Q = (.200kg)(910 J/kg-C)(80.0C)
Q = 14560 J
b)
m = 100mL = 100g = .100kg
c = 4190 J/kg-C)
T = 20.0 deg C
Q = (.100kg)(4190 J/kg-C)(20.0C)
Q = 8380 J
Ok, now this is where I'm confused. I don't know how to incorporate both of the temperatures when they are interacting with each other, or even what I did above was necessary.
Any suggestions? Thanks