Posted by bon on Tuesday, January 23, 2007 at 1:55pm.
When 80.0 grams of a certain metal at 90.0 °C was mixed with 100.0 grams of water at 30.0 °C, the final equilibrium temperature of the mixture was 36.0 °C. What is the specific heat (J/g°C) of the metal?
I suppose I would use this formula:
q= mcÄt
m= mass(g)
q= enthalpy change (j)
c= specifice heat capacity(J/g°C)
Ät= temperature change(°C)
not quite sure how to approach this question!?!?!?!
The sum of the heats added is zero (one is lost, so it is negative).
Heataddedwater + heataddedmetal=0
mw*cw*(TfTiw) + mm*Cm*(TfTim)=0
solve for Cm, the specific heat of the metal.

chem+ help!  jacqueline, Saturday, February 6, 2016 at 2:09pm
0.581J/g c
Answer This Question
Related Questions
 Physics  400 grams of a metal at 300 C is added to 100 grams of of water at 20 ...
 physics  What is the final equilibrium temperature when 40.0 grams of ice at 0°...
 Science  A piece of unknown metal with a mass of 68.6 grams is heated to an ...
 chemistry  if 50 grams of a substance with temperature of 20 degrees C and a ...
 chemistry  Twenty grams of aluminum metal at 83.2 C were mixed with 49.3 g of ...
 Physics  One hundred twenty grams of metal at 88 degree Celsius is poured into ...
 physics  A 200 gram copper calorimeter contains 400 grams of water at 25 C. A ...
 Chemistry  A coffee cup calorimeter contains 25.0 grams water at 23.8 C A 5.00g...
 Physics  120 grams of metal at 88 degree celcius is poured into a 70 grams ...
 chemistry  A 115 g piece of metal, initially at 60.0 °C, is submerged into 100....
More Related Questions