chazgurl4life
What will be the equilibrium temperature when a 277 g block of copper at 321°C is placed in a 138 g aluminum calorimeter cup containing 829 g of water at 11.6° c?

OK SO IM USING THIS EQUATION

MASS(COPPER)*SPECIFICHEAT(COPPER)*(initaltempf copper-Teq)=MASS(ALUMINUM)*SPECIFICHEAT(ALUMINUM)*(Teq-initaltempof aluminum) + MASS(WATER)SPECIFICHEAT(WATER)*(Teq-initaltemp OF WATER)

WHICH ALL COMES OUT TO:

.277KG(390J/KG)(321 C-Teq)={(.138 KG)(900 J/KG) +(.825KG)(4186 J/KG)}*Teq-11.6 DEGREES CELSIUS

BUT I KEEP GETTING 1.9 DEGREES CELSIUS AS MY ANSWER WHICH IS WRONG PLEASE HELP
Thomas_L
my physics teacher explained this problem to me... I didnt get it XD could you take the formula:

T_f = (mx)(cx)(Tix)+(my)(cy)(Tiy) over (mx)(cx)+(my)(cy)

start with the cup and water then the result of that with the burner?

I dont know if this works... never really tried but hell if it helps then alls cool... blah now I wish I had taken more thourogh notes...
Drude
QUOTE
What will be the equilibrium temperature when a 277 g block of copper at 321°C is placed in a 138 g aluminum calorimeter cup containing 829 g of water at 11.6° c?

Of course in a system that is isolated the heat transfer is symmetrical with equal amounts of heat lost and gained by the parties that lose heat and those that gain.

however you have to keep in mind that although the cooper block cools it would also heat up the water and water just might boil! and evaporate

but how can you see that ?

you have to measure the heat contained in the cooper block , and measure the amoutn heat needed to raise the temperature of water to 100 and also the heat of fusion Q=mL and if the heat content of cooper is more than the heat of fusion + heat of water then you have to modify the equation.
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