Brandtley12
Answered

Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

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.

Sagot :

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°
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.