laboratory technician drops a 0.0850kg sample of unknown solidby Larrytsai Tags: drops, laboratory, sample, solid, technician, unknown 

#1
Sep2009, 05:16 PM

P: 228

A laboratory technician drops a 0.0850kg sample of unknown solid material, at a temperature of 100.0 degree celcius, into a calorimeter. The calorimeter can, initially at 19.0 degrees celcius, is made of 0.150kg of copper and contains 0.200kg of water. The final temperature of the calorimeter can and contents is 26.1 degrees celcius. Compute the specific heat of the sample.
The only thing i have so far is Qsys= Qsurr but im stumped with the 2 specific heat variables 



#2
Sep2109, 02:36 AM

Admin
P: 22,712

Not sys and surr, I would rather go for gain=lost.
I suppose you should check both specific heats of copper and water in tables.  



#3
Sep2509, 02:32 PM

HW Helper
P: 1,933

Being calories and degrees C he ought to know that for water roughly.




#4
Oct2209, 03:29 AM

P: 8

laboratory technician drops a 0.0850kg sample of unknown solida 30.14g stainless steel ball bearing at 117.82 c is placed in a constantpressure calorimeter containing 120.0 mL of water at 18.44 C.if the specific heat of the ball bearing is 0.474 J/g.c, calculate the final temperature of the water.assume the calorimeter to have negligible capaity. 



#5
Oct2209, 03:38 AM

Admin
P: 22,712

It is again simple heat gain (by water) equals heat lost (by ball). Assume final temperature to be T_{final} and write equation for a heat balance. You will get equation with one unknown. That's all.
 methods 



#6
Oct2409, 01:32 AM

P: 8

prove that (Delta Temp) for water equals (Delata Temp) for the ball!!!!! 



#7
Oct2409, 01:46 AM

P: 274

No, but both changes in temperature are directly proportional to the amount of heat transferred.



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