Question

An object with a mass of `20 g` is dropped into `120 mL` of water at `0^@C`. If the object cools by `60 ^@C` and the water warms by `3 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`1260 "J kg"^{-1}"K"^{-1}`

Explanation:

The specific heat capacity of water is `4200 "J kg"^{-1}"K"^{-1}`. So, the heat capacity of 120 mL = 0.12 kg water is `504 "J K"^{-1}`. Thus the heat gained by the water is 1512 J. Assuming that no energy is lost in the process, this is the heat lost by the body in cooling down by 60 K. Thus the specific heat capacity of the material is given by

`{1512 "J"}/{0.02 "kg" xx 60 "K"} = 1260 "J kg"^{-1}"K"^{-1}`