HW #3 Problem: An aluminum cup contains 225 g of water
at 27°C. A 400g sample of silver at an
initial temperature of 56°C is placed in the water. A 40 g copper
stirrer, that starts in
the water, is used to stir the mixture until the entire system
reaches a final
temperature of 29°C. What is the mass of the aluminum
cup?
Given mass of water mw=225g,
Mass of silver mAg=400g, Mass of copper mcu=40g
Mass of aluminum cup mal=????
Also initial temperature of water and copper is Tw=Tcu=27°C
Initial temperature of silver is TAg=56°c
Now to find out the mass of aluminum cup we use conservation of heat energy.
That is Q of water+copper+aluminum=Qof silver
So as we know Q=mcΔT, where m=mass, c= specific heat, ΔT=change in temperature
Now total heat energy of aluminum+copper+water=Q
Therefore Q=(mAlCAl+mwCw+mcuCcu)ΔT
Where C of Al=0.900, C of copper=0.386, C of water=4.186
Therefore Q=[(mAl×900+225×4186+40×386)(29-27)]/1000
Q=[mAl×900+941850+15440](2)/1000
Now QAg(of silver)=[400×233×(56-29)]/1000
Now according to conservation of energy these two energy should be equal.
Therefore Q=QAg
[mAl×900+941850+15440](2)/1000=[400×233×27]/1000
900mAl=(2516400/2)-(941850+15440)
mAl=334.34g
HW #3 Problem: An aluminum cup contains 225 g of water at 27°C. A 400g sample...
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