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
Feb 3, 2017

Answer:

#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)#