Question

A 0.5kg piece of metal (c = 600/kgK) at 300 degree celcius is dumped into a large pool of water at 20 degrees celcius. Assuming the change in temperature of water to be negligible, calculate the overall change in entropy for the system

Answer 1

we can assume that the change in temperature of water is meanless (because it is a large pool of water), so the final temperature is 20 dgree celsius.
so, \[\Delta T= (300-20)K\]
Then, you can determine value of Q with equation \[Q= m*c*\Delta T\]
Then \[\Delta S=Q/T\]

I'm stuck in here, i don't know, we must choose T in wich?

May be, my brother @quarkine and @RaphaelFilgueiras can explain it anymore.

Answer 2

@Rizki_Zhou pls what formula is this ΔS=Q/T

Answer 3

\[\Delta S= value-of-entropy-change\]

Answer 4

you can see it in http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node41.html

Answer 5

Pls guys i need help is @Rizki_Zhou in the right direction? and what is the value of T

Answer 6

\[dS=dQ/T\]
\[dS=d(m*c*dT)/T\]
\[dS=m*c*dT/T\]
\[DeltaS=\int\limits_{?}^{?}m*c*dT/T\]
\[DeltaS=m*c*\int\limits_{?}^{?}dT/T\]
ΔS=m*c*ln Tf/Ti

that should be entropy change for the piece of metal.
for the system i am not sure........

Answer 7

@mos1635 wat is T final and T initial since we assume temp change is not significant

Answer 8

we assume temp change is not significant for the water pool
not for the piece of metal

Answer 9

i think you made mathematical error.

Answer 10

T final is same as that of the cold body and T initial is that of the metal before dumping...since the temp change for cold reservoir is negligible i.e its dT=0..hence entropy change is positive and it is m*c*deltaT()/T(cold) i.e negligibly small...@experimentX :what mistake did he do??

Answer 11

\( \int 1/T dt = ln |T| + C\)

Answer 12

integral is from ti to tf so it came out as ln (Tf/Ti)..didnt it??

Answer 13

well, mathematically it should be in that way.

Answer 14

\[\int\limits_{Ti}^{Tf?}(1/T)dT=\ln T (from Ti \to Tf)\]
lnTf-lnTi=ln(Tf/Ti)

Answer 15

also integrating gives the total entropy ... not change in entropy.

Answer 16

@all im lost here..wat the final solution to this

Answer 17

i think it is true as the hot body is getting colder so its entropy change is negative...since ti is more than tf,this should work fine...also we integrate both sides so there is a Sf-Si term on left side which is same as change in entropy...

Answer 18

@woleraymond you know the answer??

Answer 19

also integrating gives the total entropy ... not change in entropy.
on the contrary my dear Watson on the contrary
Q is the heat during a change not for a state
so ΔS it is

Answer 20

please check heat transfer between tow reservoirs
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node41.html

Answer 21

ΔS for pool
ΔS=Qfrom metal/T of pool
ΔS=m*C*|ΔT| of metal/T of pool
and finaly
ΔS total=ΔSmetal+ΔSpool

Answer 22

also deltaT()/ i wrote is actually Tf-Ti itself andit wont be negligile..
@experimentX :that link which @Rizki_Zhou shared is a nice one but the thing is the metal piece is not a reservoir...

Answer 23

do we really have to do |ΔT| for metal??

Answer 24

oh yeah, google had answer all our question.

Answer 25

give them then to @woleraymond

Answer 26

@experimentX of course not...else we will lose the info about whether entropy is increasing or decreasing...

Answer 27

@experimentX i dont know d answer...@Rizki_Zhou pls what is the solution

Answer 28

not really sure if this works out @mos1635 was right about earlier.
http://www.wolframalpha.com/input/?i=0.5+*+600+*+ln+%28293%2F573%29+%2B+0.5+*+600+*+280%2F293

Answer 29

@experimentX pls clarify this equation 0.5 * 600 * ln (293/573) + 0.5 * 600 * 280/293

how do u come about 280

i also tink @mos1635 was right too

Answer 30

0.5 * 600 * ln (293/573) <--- entropy change of metal piece
0.5 * 600 * 280/293 <----- entropy of sink

Answer 31

first one is same as of mos
the second is delQ/T provided in the link pasted above.

Answer 32

I think i got around 85.5

Answer 33

looks like i didn't see ln in mos equation ... sorry about that

Answer 34

@experimentX 280 is 300- 20 in degrees why others are in kelvin, why?

Answer 35

http://www.assignmentexpert.com/free-questions/question-on-physics-molecular-physics-and-thermodynamics-8649.html

looks like it's right
the difference in kelvin scale and Celsius scale is same.

Answer 36

thanks @experimentX

Answer 37

thanks to everyone