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how much thermal energy is LOST while cooling a 10g block of aluminum form 120 celcius to 80 celcious

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\[\Delta Q = m*c*\Delta T\]
locate c, the specific heat of aluminum, on a table... use units for mass that agree with the units of the value you have for specific heat...
is the specific heat for aluminum 0.920 J/(g c)? is that what ur asking for?

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Other answers:

do you want me to google that in order to verify that it's the correct value or something?
ok so i take that and i multiply it by the mass which is 10g but how do i find the delta t?
from 120 to 80 is a change of?
and if u want u can google it i have it on my homework worksheet though
40 degrees
i think so
if you had 6 twenties in your wallet and then you checked again in a few minutes and you only had 4 twenties, what happened?
you lost 40 dollars
120 -> 80 is a change of -40
so it would be negative 40?
ok so the formula would be 5*0.920J*-40 then i would get my answer?
it's 10 grams
oh sorry i was looking at the wrong problem lol
so its -368 J
sure but the question asks how much heat is lost... so if I asked you how much money you lost in the earlier example... what would you say?
-40 dollars? or 40 dollars?
negative 40
"I lost negative forty dollars" does anyone ever say that?
"lost" means "negative".
no lol so we would say we lost 40 thermal energy?
temperature decreased -> heat was lost.... delta T is negative delta Q is negative delta Q = -386 J means 386 J was lost
oooh ok thank you!!!

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