An object with a mass of #4 g# is dropped into #250 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #2 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Feb 16, 2017

Answer:

The specific heat of the material is #=1,042 Kcalkg^-1°C^-1#

Explanation:

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

For the cold water, # Delta T_w=2ºC#

For the object #DeltaT_o=120ºC#

# M_o C_o (DeltaT_o) = Mw C_w (Delta_w)#

#C_w=1Kcalkg^-1ºC^-1#

#0,004 C_o120 = 0,250* 1 *2#

#C_o = 0.5/ 0.48 = 1,042 Kcalkg^-1°C^-1#