"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?"(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Specific Heat of aluminum Problem

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