An object with a mass of #45 g# is dropped into #600 mL# of water at #0^@C#. If the object cools by #24 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?

Question

1389 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-14T12:16:34

The specific heat is #=18.60kJkg^-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=24ºC#

# 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*24=0.6*4.186*8#

#C_o=(0.6*4.186*8)/(0.045*24)#

#=18.60kJkg^-1K^-1#