Question

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

Answer 1

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.

Answer 2

is the answer .9

Answer 3

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.

Answer 4

THANKSS DOC.

Answer 5

You're welcome.