Question

An object with a mass of `30 g` is dropped into `50 mL` of water at `0^@C`. If the object cools by `20 ^@C` and the water warms by `80 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`cp_o = 6,66 (Kcal)/(kg xx °C) = 27,9 (KJ)/(Kg xx K)`

Explanation:

the heat gained from water (50 mL = 50 g) is the same than the heat given from the objec. Hence `M_o xx cp_o xx (Ti-Tf)_o = M_w xx cp_w xx (Tf-Ti)_w`
`0,030 g xx cp_o xx 20°C = 0,050 kg xx 1( Kcal)/(kg xx °C) xx 80°C`
`0,6 g xx °C xx cp_o = 4kcal`
`cp_o = 6,66 (Kcal)/(kg xx °C) = 27,9 (KJ)/(Kg xx K)`
Pay attention! It doesn't exist any solid object whose specific heat is bigger than water's one
Moreover sperimentally you should considered the effect of the container