Question

An object with a mass of `70 g` is dropped into `600 mL` of water at `0^@C`. If the object cools by `24 ^@C` and the water warms by `8 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`Cp_o = 2,85(Kcal)/(kg°C)`?

Explanation:

the amount of heat changed by the object (Qo) is the same changed by water(Qw).
`M_o Cp_o (T2-T1)_o = M_w Cp_w (T2-T1)_w`
since 600 mL of water are 0,6 Kg and `Cp_w` is 1 Kcal/(kg°C)
`0,07kg Cp_o 24°C = 0,6kg 1(Kcal)/(kg°C) 8°C`
`Cp_o = 2,85(Kcal)/(kg°C)`
the result is impossible because dosen't exist a solid object with a specific heat so high