An object with a mass of #110 g# is dropped into #750 mL# of water at #0^@C#. If the object cools by #70 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Scott F.
Mar 27, 2017

#C_"object"=1.22"J/g°C"#

Explanation:

This question is based on the concept that #"heat lost"="heat gained"#. That is, all the heat lost by the object is gained by the water (#q_"water"=q_"object"#).

The equation to find the change in heat is #q=m*C*DeltaT#, and #C_"water"=4.184"J/g°C"#.

Since no density or mass for water is given, it can be assumed that #rho_"water"=1"g/mL"# so there are #750"g"# of water.

Using the change in heat equation, #q_"water"=m_"water"*C_"water"*DeltaT_"water"#
#q_"object"=m_"object"*C_"object"*DeltaT_"object"#

Since #q_"water"=q_"object"#, these two equations can be set equal to each other.
#m_"water"*C_"water"*DeltaT_"water"=m_"object"*C_"object"*DeltaT_"object"#

Solving this for the desired variable,
#C_"object"=(m_"water"*C_"water"*DeltaT_"water")/(m_"object"*DeltaT_"object")#

And plugging in the numbers
#C_"object"=((750"g")(4.184"J/g°C")(3"°C"))/((110"g")(70"°C"))#

Using a calculator
#C_"object"=1.22"J/g°C"#

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