Question

a silver ball weighing 50g is removed from a furnace and dropped into a 350g of water at 0 degree celcius. if the equilibrium temperature is 10 degree celcius and the specific heat capacity of platinum is 230Jkg-1 degree celcius-1, what is the temperature of the furnace? neglect the effect of the mass of the calorimeter...

Answer 1

anyone can help me??

Answer 2

\[Q=mCv DeltaT\]
If Qw=Qs

Answer 3

\[Ms Cs (10-Tf)=Mw Cw 10\]

Answer 4

Ms mass silver
Mw mass water
Cs and Cw specific heat capacity of the silver and Water
Tf Temperature of the furnace

Answer 5

10 means what?? ten??

Answer 6

sorry is 283 K

Answer 7

sorry about that

Answer 8

283 K is the equillibrium temperature

Answer 9

indeed it is ten

Answer 10

(0.05)(230)(10-Tf)=(0.35)(4200)(10)
??

Answer 11

ok

Answer 12

i get -1268.26

Answer 13

The question first says the ball is made of silver then it later gives the specific heat of platinum to be 230J/kgC...which is actually the specific heat of silver. At any rate, I'll use the value the question provides for the ball.

The only formula you need for this problem is:
\[Q=mC \Delta T\]
First, calculate the energy required to raise the temperature of the water by 10 degrees C:
\[Q_{water}=(350g)(4.186J/gC)(10)=14.651kJ\]
For this problem, we can assume all of this energy came from the ball:
\[Q_{ball}=Q_{water}=14.651kJ\]
Now calculate the change in temperature of the ball:
\[Q_{ball}=(50g)(0.23J/gC)(\Delta T)=14.651kJ\]\[\Delta T = 1274C=T_{initial}-T_{final}\]And we know Tfinal is 10C, so that becomes:
\[1274C+10C=T_{initial}=1284C\]Therefore, the temperature of the furnace must have been 1284C.

Answer 14

thanks you...now i understand this question...;)

Answer 15

no problem :)

Answer 16

thanks a lot...