An object with a mass of #180 g# is dropped into #660 mL# of water at #0^@C#. If the object cools by #72 ^@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 · 2017-07-18T14:07:48

The specific heat is #=0.64kJkg^-1K^-1#

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

For the cold water, # Delta T_w=3^@C#

For the object #DeltaT_o=72^@C#

# 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

Mass of the object is #m_0=0.180kg#

Mass of the water is #m_w=0.66kg#

#0.18*C_o*72=0.66*4.186*3#

#C_o=(0.66*4.186*3)/(0.18*72)#

#=0.64kJkg^-1K^-1#