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?

1 Answer
Jane
Jan 26, 2018

Mass of 210 mL of water is 210 cm^3 × 1 (gm)/(cm^3) i.e 210 gm

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 gm 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 gm^-1 degree C ^-1

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