An object with a mass of 32 g is dropped into 210 mL of water at 0^@C. If the object cools by 96 ^@C and the water warms by 7 ^@C, what is the specific heat of the material that the object is made of?

Mass of $210 m L$ of water is 210 cm^3 × 1 (gm)/(cm^3) i.e $210 g m$
So, heat energy absorbed by that much water to increase its temperature by $7$ degree C will be (210×1×7) Calorie or $14700$ Calorie (using H = m×s×d theta)
And that amount of energy must have been released by that $32 g m$ of object which has cooled by $96$ degree C,whose value is (32×s×96) Calorie where $s$ is the specific heat of that object.
So, equating both we get, $s = 4.78$ Calorie $g {m}^{-} 1 \mathrm{de} g r e e {C}^{-} 1$