Posted by **Brandi** on Friday, March 1, 2013 at 8:27am.

A 115 g piece of metal, initially at 60.0 °C, is submerged into 100.0 g of water at 25.0 °C. At thermal equilibrium, this mixture had a final temperature of 27.0 °C. The specific heat capacity of water is 4.18 (J/g°C). Use this information to determine the specific heat capacity of the metal (calculate your answer to 3 sig figs).

- chemistry -
**Devron**, Friday, March 1, 2013 at 9:23am
The heat change for both objects will be equal, so you need two equations and they need to be set to each other. The equation to use is the following:

q=mcT∆

Conditions for 1

m1=mass1=115g

c1=specific heat1=?

T∆1=25.0°C-27.0ºC=-2.0Cº

Conditions for 2:

m2=mass2=100.0g

c2=specific heat2=4.18 (J/g°C)

T∆2=25.0°C-60.ºC=-30.0ºC

m1c1∆T1=m2c2∆T2

(115g)c1(-2.0Cº)=(100.0g)(4.18 (J/g°C))(-30.0ºC)

Solve for c1,

c1=[(100.0g)(4.18 (J/g°C))(-30.0ºC)]/[(115g)(-2.0Cº)]

c1=(-1.25 x 10^4 J/)(-2.3 x 10^2 gºC)

c1=54.3 J/g°C

****I think; I havent had to do this type of calculations in years.

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