A 120 Kg Piece of Granite [C_p= 0 $\degree$ C)] at 600

A 1.20 kg piece of granite [c_p= 0.790 J/(g . $\degree$ C) ] at 60.0 $\degree$ C must be cooled to 24.0 $\degree$ C by submersion in water.How much water [c_p = 4.18 J/(g . $\degree$ C) ], initially at 18 $\degree$ C, would be required?

A) 0.302 kg
B) 0.907 kg
C) 1.36 kg
D) 1.72 kg
E) 8.16 kg