A 333.33g sample of metal at 70.2 degrees C was placed in 100.43g OF water at 21.4 degrees C. The final temp. of the water and metal was 32.6 degree C. If not heat was lost the surroundings what is the specific heat capacity of the metal?

What i know: q=m*c*delta T.
but….. my teacher said we cannot add masses together only volumes, so I'm not sure what to do.

Question

A 333.33g sample of metal at 70.2 degrees C was placed in 100.43g OF water at 21.4 degrees C. The final temp. of the water and metal was 32.6 degree C. If not heat was lost the surroundings what is the specific heat capacity of the metal?

What i know: q=m*c*delta T.
but….. my teacher said we cannot add masses together only volumes, so I'm not sure what to do.

aaronq · Answer

you use that equation.. twice. The heat given off by the metal is absorbed by the water, in mathematical terms this is:  $-q_{metal}=q_{water}$

so you do just that,   $-q_{metal}=-[m*C_p*\Delta T]=m*C_p*\Delta T=q_{water}$

jfraser · Answer

in the equation that @aaronq shows, the only value you don't know is the $C_p$ of the metal. You don't have to directly solve for the heat transferred, but you can if you want to break the problem into smaller pieces

anonymous · Answer

I found that the specific heat captivity of the metal is 0.126, am I right?

anonymous · Answer

Capacity*

jfraser · Answer

without doing the calculation myself, it sounds about right. metals have very low specific heat capacity, it's what makes themsuch good conductors of heat and electricity. water has a specific heat capacity of 4.18 J/gC, so a metal with a C of 0.126 is very reasonable

vincent-lyon.fr · Answer

Hi! 
I find:    $c_P$ = 0.375 J/g/°C 

which looks like the metal was copper.
Try to plug in your values again.