An object with a mass of #10 g# is dropped into #120 mL# of water at #0^@C#. If the object cools by #60 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-06-02T17:51:39

The specific heat is #=2.51kJkg^-1K^-1#

The heat is transferred from the hot object to the cold water.

For the cold water, # Delta T_w=3ºC#

For the object #DeltaT_o=60ºC#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

#C_w=4.186kJkg^-1K^-1#

Let, #C_o# be the specific heat of the object

#0.010*C_o*60=0.12*4.186*3#

#C_o=(0.12*4.186*3)/(0.010*60)#

#=2.51kJkg^-1K^-1#