An object with a mass of #60 g# is dropped into #700 mL# of water at #0^@C#. If the object cools by #40 ^@C# and the water warms by #16 ^@C#, what is the specific heat of the material that the object is made of?

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Answer 1 · 2016-02-17T18:09:27

#19553.33J//kg.K#

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

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

#therefore (0.700)(4190)(16)=(0.060)(c)(40)#

#therefore c=19553.33J//kg.K#