An object with a mass of #45 g# is dropped into #650 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?

Question

1435 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-11T12:58:28

The specific heat is #=4.03kJkg^-1K^-1#

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

For the cold water, # Delta T_w=8ºC#

For the object #DeltaT_o=120ºC#

# 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

#0.045*C_o*120=0.65*4.186*8#

#C_o=(0.65*4.186*8)/(0.045*120)#

#=4.03kJkg^-1K^-1#