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.

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°