# An object with a mass of 120 g is dropped into 640 mL of water at 0^@C. If the object cools by 36 ^@C and the water warms by 12 ^@C, what is the specific heat of the material that the object is made of?

Jan 23, 2018

To come in thermal equilibrium,heat energy released by the object is m×s×del theta = 120×s×36  Calorie.(where,$s$ is its specific heat)
And during this process,heat taken by water is 640×1×1×12 Calorie (Here,mass of water = volume×density)
So,equating both we get, $s = 1.77$ Calorie gm^-1 'C^-1