An object with a mass of #32 g# is dropped into #210 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?

Question

1288 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-14T07:28:14

The specific heat is #=1.37kJkg^-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

#0.032*C_o*60=0.210*4.186*3#

#C_o=(0.210*4.186*3)/(0.032*60)#

#=1.37kJkg^-1K^-1#