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Reps for chemistry question help!!!!! (srs)
How much heat energy is required to raise the temperature of 0.359 kg of copper from 23.0 C to 60.0 C If the specific heat of copper is 0.0920 cal/(g x C)
That is a times sign in parenthesis and C stands for degrees celsius, I cant figure out how to do superscript on here.

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Registered User
Q = mcΔT
Q is heat, just plug in and solve.
also convert your kg's to g

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Originally Posted by CurlsfotheGirls
is this srs?
do you have no formulas?
even still you should be able to solve this.
I am horrible at chemistry man, retaking this course AGAIN
formula is SH= Heat
___________
Mass x Delta T (temperature change)

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Originally Posted by UnflappedAcorn
not going through brah.

Proudest manlet
1222? i forget the damn formula

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Originally Posted by CurlsfotheGirls
this
and check your units.
if you need to; make sure to convert from cal to Joules.
edit: if your not gettting it post brah and i got you
so the Kg to G is 359, Joules is fuaaaaaaaaaaaaark!!!!!!1 I cant get this **** man.

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Originally Posted by Shreder54
1.22? i forget the damn formula
Forgot to convert kilograms to grams

Are your acorns flapped?
Originally Posted by lift_heavy20
maybe its the significant units
the exact is: 13.283

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Originally Posted by lift_heavy20
so the Kg to G is 359, Joules is fuaaaaaaaaaaaaark!!!!!!1 I cant get this **** man.
1 cal = 4.186 J I think
I haven't taken chem in like 3 years, this is stuff I remember from physics

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Originally Posted by captiontom
Forgot to convert kilograms to grams
Yeah you're right, i thought I did. It's 1222 then I think.
It's really easy if you look in book or look it up online...

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Q = mcΔT
Q = 359g*0.0920*37
= 1222.036 cal
1 cal = 4.184 J
1222.036cal = 5112.998J

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thanks guys it was actually 1222, anyone want to explain how this was achieved?

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I think I repped everyone in here thanks guys, plz explain how answer was achieved?

I'm so money.
Originally Posted by lift_heavy20
I think I repped everyone in here thanks guys, plz explain how answer was achieved?
there is a formula...q=mc(delta)T
m is the mass of the copper
c is the specific heat
deltaT is the change in temp (in kelvin= C + 273)
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Originally Posted by lift_heavy20
thanks guys it was actually 1222, anyone want to explain how this was achieved?
you don't need to do any cal to J conversions.
just convert your kg to g so you get 359g.
Q = mcΔt
m=359 g
c is the specific heat, in this case it's copper
Δt is the change in temperature so it's 6023
just plug that all in and you should get 1222

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Originally Posted by lift_heavy20
thanks guys it was actually 1222, anyone want to explain how this was achieved?
First convert kilograms of copper to grams
Second determine the change in temperature
Third multiply the specific heat by the grams of copper by the change in temperature
The answer is in calories

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it's just a basic formula equation bro, don't you know the formula?
q= mcT
= 35.9g*0.092*37C
= ....
had an chem exam this monday and this was one of the chapters

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Originally Posted by thercias
it's just a basic formula equation bro, don't you know the formula?
q= mcT
= 35.9g*0.092*37C
= ....
had an chem exam this monday and this was one of the chapters
And you missed this problem...
359 grams not 35.9

Banned
I love working with copper at work * plumber *

watchout ur comments bro^
Originally Posted by Leh
Q = mcΔT
Q = 359g*0.0920*37
= 1222.036 cal
1 cal = 4.184 J
1222.036cal = 5112.998J
somebody repp this mawfukkah right now
EDIT: Strong first post brah
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