An object with a mass of #15 g# is dropped into #250 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #9 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Apr 30, 2018

From your data we can assume the object lost heat and the water gained heat in equilibrating.

Moreover, I am assuming:

#rho_(H_2O) = (1.0"g")/"mL"#

Recall,

#q = mC_sDeltaT#

In this case, the heat gained by the water is supplied by the body. Accordingly, we can rearrange the equation, such that,

#-m_xC_xDeltaT = m_(H_2O)C_(H_2O)DeltaT_(H_2O)#

Hence,

#=> C_x = (-m_(H_2O)C_(H_2O)DeltaT_(H_2O))/(mxDeltaT_x) approx (5.23"J")/("g"*°"C")#

is the specific heat of the body of mass we dropped into the water.