An object with a mass of #32 g# is dropped into #480 mL# of water at #0^@C#. If the object cools by #80 ^@C# and the water warms by #4 ^@C#, what is the specific heat of the material that the object is made of?

Question

1085 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-18T04:59:29

The specific heat is #=3.14kJkg^-1K^-1#

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

For the cold water, # Delta T_w=4ºC#

For the object #DeltaT_o=80ºC#

# 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*80=0.48*4.186*4#

#C_o=(0.48*4.186*4)/(0.032*80)#

#=3.14kJkg^-1K^-1#