110.0 gm * sh * (80°C - 24.0°C) = ...
... 120.0 gm * 4.186 J/gm⋅°C * (24.0°C - 21.0°C)
What is the specific heat capacity of a
110.0 gram piece of metal initially at 80°C that
will change 120.0g of 21.0°C water to 24.0°C?
2 answers
[mass metal x specific heat metal x (Tfinal - Tinitial)] + [mass H2O x specific heat H2O x (Tfinal - Tinitial) = 0
You have mass metal, Tf metal is 24 C, Ti for metal is 80 C, mass H2O is given, specific heat H2O you have in your notes or text book or on the web, Tfinal H2O is 24 C and Tinitial H2O is 21.0 C.
You have mass metal, Tf metal is 24 C, Ti for metal is 80 C, mass H2O is given, specific heat H2O you have in your notes or text book or on the web, Tfinal H2O is 24 C and Tinitial H2O is 21.0 C.
0 answers