An object with a mass of #90 g# is dropped into #500 mL# of water at #0^@C#. If the object cools by #60 ^@C# and the water warms by #5 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Apr 24, 2017

Answer:

The specific heat is #=1.94kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=5ºC#

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

#0.090*C_o*60=0.5*4.186*5#

#C_o=(0.5*4.186*5)/(0.090*60)#

#=1.94kJkg^-1K^-1#