faby9898
how much thermal energy is LOST while cooling a 10g block of aluminum form 120 celcius to 80 celcious



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Algebraic!
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\[\Delta Q = m*c*\Delta T\]

Algebraic!
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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...

faby9898
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is the specific heat for aluminum 0.920 J/(g c)? is that what ur asking for?

Algebraic!
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sure.

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do you want me to google that in order to verify that it's the correct value or something?

faby9898
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ok so i take that and i multiply it by the mass which is 10g but how do i find the delta t?

Algebraic!
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from 120 to 80 is a change of?

faby9898
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and if u want u can google it i have it on my homework worksheet though

faby9898
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40 degrees

Algebraic!
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+40?

faby9898
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i think so

Algebraic!
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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?

Algebraic!
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you lost 40 dollars

Algebraic!
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120 > 80 is a change of 40

faby9898
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so it would be negative 40?

Algebraic!
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yes:)

faby9898
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ok so the formula would be 5*0.920J*40 then i would get my answer?

Algebraic!
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it's 10 grams

faby9898
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oh sorry i was looking at the
wrong problem lol

faby9898
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so its 368 J

Algebraic!
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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?

Algebraic!
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40 dollars? or 40 dollars?

faby9898
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negative 40

Algebraic!
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"I lost negative forty dollars"
does anyone ever say that?

Algebraic!
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"lost" means "negative".

faby9898
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no lol so we would say we lost 40 thermal energy?

Algebraic!
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temperature decreased > heat was lost....
delta T is negative delta Q is negative
delta Q = 386 J means 386 J was lost

faby9898
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oooh ok thank you!!!