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?

1 Answer
Andrea B.
Feb 3, 2017

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)

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