As 390-g of hot milk cools in a mug, it transfers 30,000 J of heat to the environment.
How does the temperature of the milk change? (cmilk = 3.9 J/g°C)
It decreases by 77°C.
It decreases by 20°C.
It increases by 77°C.
It increases by 20°C
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OpenStudy (andrewkaiser333):
@anyone
OpenStudy (andrewkaiser333):
@hartnn
OpenStudy (andrewkaiser333):
@pig_cat223
OpenStudy (pig_cat223):
hang on and ill try i gotta problem of my own...
OpenStudy (andrewkaiser333):
ok
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OpenStudy (theprofessor35):
Hmmm
OpenStudy (andrewkaiser333):
anyone know??
OpenStudy (pig_cat223):
im sorry but i have no idea what is, i dont even know where to start and i cant find anything online either.... sorry and thanks for the help on mine... :D
OpenStudy (andrewkaiser333):
np
OpenStudy (pig_cat223):
:/
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OpenStudy (andrewkaiser333):
know of anyone who can help???
OpenStudy (pig_cat223):
i tried asking for you in the math chat box but they were typing to fast to see it.... but you should ask
OpenStudy (andrewkaiser333):
ok
OpenStudy (andrewkaiser333):
@Episode
OpenStudy (episode):
We can apply Newton's Law of cooling yes?
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OpenStudy (andrewkaiser333):
yes?
OpenStudy (andrewkaiser333):
I am not that good at this
OpenStudy (episode):
Do you know calculus?
OpenStudy (andrewkaiser333):
not that well
OpenStudy (andrewkaiser333):
been 3 years since i did it last so very rusty
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OpenStudy (episode):
This is specific heat
OpenStudy (episode):
So I guess we don't really need calculus we can apply the formula you had earlier, \[Q = c m \Delta T \implies \Delta T = \frac{ Q }{ cm }\] since delta t is the change in temperature
OpenStudy (episode):
That will give you your result :)
OpenStudy (andrewkaiser333):
ok
OpenStudy (episode):
remember c is the specific heat, and m is the mass
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OpenStudy (episode):
Q is change in temp
OpenStudy (andrewkaiser333):
so can you help me plug it in
OpenStudy (episode):
Try plugging it in yourself and I will tell you if you're right or not
OpenStudy (andrewkaiser333):
thing is not sure what is what
OpenStudy (andrewkaiser333):
this is why i am asking for u to help please
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OpenStudy (episode):
You have the formula and variables read your question carefully and try, and I will show you the right answer after
OpenStudy (andrewkaiser333):
^T=30000/390(30000) ????
OpenStudy (andrewkaiser333):
Like i said I am not good with this i think it got this wrong
OpenStudy (episode):
Careful, \[\Delta T = \frac{ 30,000 J }{ (3.9)(390g)}\] If you plug in all the units you will see, you get the temperature
OpenStudy (episode):
You almost had it right, you put Q again instead of the specific heat
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OpenStudy (andrewkaiser333):
\[DeltaT=30000/(3.9)(390)=3,000,000\]
OpenStudy (andrewkaiser333):
am i right?
OpenStudy (episode):
Nope, plug it in your calculator
OpenStudy (andrewkaiser333):
hold on
OpenStudy (andrewkaiser333):
same answer what am i doing wrong
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OpenStudy (episode):
You're dividing right? 30,000J by 3.9*390
OpenStudy (andrewkaiser333):
yes
OpenStudy (episode):
Then you should get 19.7 degrees
OpenStudy (photon336):
hey @AndrewKaiser333
\[q = m~C~\Delta T \]
OpenStudy (andrewkaiser333):
i see what i did now. thanks
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OpenStudy (episode):
Yw :)
OpenStudy (andrewkaiser333):
so increases by 20°
OpenStudy (episode):
Right
OpenStudy (andrewkaiser333):
thanks this helps alot
OpenStudy (episode):
Yes, that is what we applied @Photon336 if you scroll up a bit
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OpenStudy (photon336):
energy is lost to the environment. q,
\[\frac{ -30,000J }{ 390G*3.9Jg^{-1}c^{-1}} = \Delta T \]
OpenStudy (photon336):
@episode heat is lost to the environment so I'm guessing that Q would be negative
OpenStudy (episode):
Yes, you are right, I did not read that part! Good catch @Photon336
OpenStudy (episode):
@AndrewKaiser333
As photon mentioned the mug would cool, so the temperature would decrease
OpenStudy (photon336):
Yeah that's one of those things I get confused by and as @Episode said that's important because that will show you how the temperature would change.
Cheers guys
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