An object with a mass of #64 g# is dropped into #560 mL# of water at #0^@C#. If the object cools by #80 ^@C# and the water warms by #6 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Feb 26, 2016

Answer:

#2749.69J//kg.K#

Explanation:

Conservation of thermal energy : Heat energy lost by object equals heat energy gained by water.

#therefore (mcDeltaT)_(object)=(mcDeltaT)_(water)#

#therefore 0.064xxcxx80=0.560xx4190xx6#

#therefore c=2749.69J//kg.K#