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

its mg hope this

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).