Question

An object with a mass of `160 g` is dropped into `750 mL` of water at `0^@C`. If the object cools by `120 ^@C` and the water warms by `5 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

`0,203 (Kcal)/(kg°C)= 0,853 (KJ)/(Kg°C)`

Explanation:

the heat transferred by the hot object, from T= To is the same than the heat absorbed by the cool water from T= Tw.
T1 is the final temperture =5°C
`Mo Cpo (To-T_1) = Mw Cpw (T_1-Tw)`
`0,160 kg Cpo (120-5) °C = 0,750 Kg (1(Kcal)/(kg°C)) (5-0)°C`
`Cpo = (3,75Kcal)/ (18,4 (Kg°C)) = 0,203 (Kcal)/(kg°C)= 0,853 (KJ)/(Kg°C)`