Asked by miley

another calorimetry experiment is performed in which a 500. gram peice of gray metal at 67 degrees celsius is added to 493 grams of water. the water's temperature, originally at 17 degrees celsius, changes to 22 degrees celsius. What is the specific heat of the gray metal? I know that you use the formula q=m*c* delta T. but how do i do this problem
Answered by DrBob222
q for metal + q for water = 0
q for metal is m*c*delta t.
q for water is m*c*delta t.
The only unknown is c for the metal.
Answered by Miley
is the answer .9
Answered by DrBob222
No it isn't. You may not be substituting correctly. For delta T, make sure it is (Tfinal-Tinitial). That will make the signs right. I found an answer less than 1 J/g*C.
Answered by Miley
THANKSS DOC.
Answered by DrBob222
You're welcome.