An object with a mass of #4 g# is dropped into #650 mL# of water at #0^@C#. If the object cools by #300 ^@C# and the water warms by #30 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Feb 18, 2017

Answer:

The specific heat of the material is #=68.0 KJkg^-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=30ºC#

For the object #DeltaT_o=300ºC#

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

#C_w=4.186KJkg^-1ºC^-1#

#0,004* C_o*300 = 0,6500* 4.186 *30#

#C_o = (0.650*4.186*30)/(0.004*300) = 68.0 KJkg^-1°C^-1#