posted by bon on .
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= 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*(Tf-Tiw) + mm*Cm*(Tf-Tim)=0
solve for Cm, the specific heat of the metal.