Brandtley12
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PLEASE HELP!!!!
You place 300g of water in a coffee-cup calorimeter and the temperature of the water is 22.3 degrees C. A 94.5g piece of silver metal is heated to 100 degrees C amd added to the water in the calorimeter. What will the final temperature of the silver and the water be? the specific heat of silver is .235 J/g degreesC. Water is 4.184 J/g degreesC.



Answer :

We just set the the changes in Energy of the two Object equal to each other:

300g * 4.184J/g * (temperature(final)C - 22.3C) =
 94.5g * 0.235 J/g *(temperature(final)C - 94.5C)

1255.2 J * (temperature(final)C - 22.3C)=
22.2075 J * (temperature(final)C - 94.5C)

1255.2 J * t(f) - 27990.96 JC=
22.2075 J * t(f) - 2098.60875JC

1232.9925 J* t(f) = 30089.56875 JC
temperature(final) = 24.4036916283 C° = 24.404 C°

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