Laboratory technician drops a 0.0850kg sample of unknown solid

[Larrytsai]

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 I am stumped with the 2 specific heat variables

 

[Borek]

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.

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[epenguin]

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

 

[traffic_boy]

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.

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ChemBuddy chemical calculators - buffer calculator, stoichiometry calculator
www.ph-meter.info - ph meter, ph electrode

i'll be so grateful if you can help me for this problem:
a 30.14-g stainless steel ball bearing at 117.82 c is placed in a constant-pressure 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.

 

[Borek]

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.

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methods

 

[traffic_boy]

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.

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chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods

thank you so much,,i got the idea, but I'm still thinking about if q(water) equals q(ball), can we
prove that (Delta Temp) for water equals (Delata Temp) for the ball!

 

[PhaseShifter]

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