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.
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
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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