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Mathematics 7 Online
OpenStudy (andrewkaiser333):

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

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

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):

:/

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?

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

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

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

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

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

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

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

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

OpenStudy (episode):

Thanks @Photon336 :)

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