An object with a mass of #120 g# is dropped into #640 mL# of water at #0^@C#. If the object cools by #6 ^@C# and the water warms by #24 ^@C#, what is the specific heat of the material that the object is made of?

Question

1322 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-23T13:54:54

The specific heat is #=89.3kJkg^-1K^-1#

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

For the cold water, # Delta T_w=24ºC#

For the object #DeltaT_o=6ºC#

# 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

The mass of the object is #m=0.120kg#

#0.120*C_o*6=0.640*4.186*24#

#C_o=(0.640*4.186*24)/(0.120*6)#

#=89.3kJkg^-1K^-1#