A276.0 g piece of granite, heated to 596.0°c in a campfire, is dropped into 1.45 l water (d = 1.00 g/ml) at 25.0°c.
the molar heat capacity of water is cp, water = 75.3 j/(mol ·°c), and the specific heat of granite is cs, granite = 0.790 j/(g ·°c).
calculate the final temperature of the granite in celsius.

 Final temperature of the granite ≈ 44.82 °C

Let T be the equilibrium temperature (in °C) to be found.

Moles of water;

(1.45 L H2O) x (1000 mL/L) x (1.00 g/mL) / (18.01 g H2O/mol)

    = 80.512 mol H2O  

Heat gained by water;

= (75.3 J/(mol ·°C)) x (80.512 mol) x (T - 25.0)°C

= (6062.465 T - 151561.6)

Heat lost by granite;

= (0.790 J/(g ·°C)) x (276 g) x (596 - T)°C

= (129951.84 - 218.04 T)

Set the two expressions for heat gained/lost equal to each other:  

6062.465 T - 151561.6= 129951.84 - 218.04 T  

6280.505 T = 281513.44

                 T = 44.8234  

                 T ≈ 44.82 °C

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