Sunday
November 23, 2014

Homework Help: AP Chem

Posted by Jess on Thursday, February 8, 2007 at 3:38pm.

I missed a lab in class a week ago and had to make it up just the other day. My teacher didn't have time to explain the lab to me, so he gave me a brief overview of the procedure. The objective of the lab was to calculate the specific heat of an unknown block of metal.

The procedure:
1. Mass metal block. Place it in beaker of water.
2. Bring beaker of water to a boil (100 degrees C).
3. Measure a known amount of water; pour this into a Styrofoam cup. Cut a second styrofoam cup so that it fits as a lid/top to the first cup. Stick a thermometer through the top of this cup. This will serve as a calorimeter.
4. Once [step 2] the water has reached a boil, wait for 2-3 minutes, then take cube out of water and place in make-shift calorimeter. Record the temperature change of the water.

The information I know is this:
Mass of block= 274.000 +/- .005 g
Amount of water in calorimeter= 130.0 +/- .5 mL
Initial temp of the block= 100 degrees C
Final temp of the block= 34.0 +/- .5 degrees C [I assume - my teacher told us to assume that the temperature of the water and the block would equilibrate, so they should be the same, correct?]
Initial temp of water in calorimeter= 23.0 +/- .5 degrees C
Final temp water= 34.0 +/- .5 degrees C

My question is how do I calculate specific heat? My idea is this, but I'm not sure if it is correct:

specific heat block= q/m*delta T
I know the delta T and m, but the q is unknown. Could I figure this out like this?

q H20= 130.0 g H20 x specific heat h20 [i'd have to look up the value] x delta-t

Would I be able to fill this value in for q to find the specific heat of the block, or no?


heat lost by block + heat gained by water = 0

massblock*sp.h.block*(Tf-Ti)+massH2O*sp.h.H2O*(Tf-Ti)=0
Tf is final T as appropriate for block and H2O.
Ti is initial T as appropriate for block and H2O.
Solve for sp.h. block.
Specific heat water = 4.18 J/g*C.

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