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*(Tf-Tiw) + mm*Cm*(Tf-Tim)=0

solve for Cm, the specific heat of the metal.

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