An object with a mass of #120 g# is dropped into #720 mL# of water at #0^@C#. If the object cools by #24 ^@C# and the water warms by #9 ^@C#, what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-04-03T13:32:35

The specific heat is #=9.42kJkg^-1K^-1#

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

For the cold water, # Delta T_w=9ºC#

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

#0.120*C_o*24=0.72*4.186*9#

#C_o=(0.72*4.186*9)/(0.120*24)#

#=9.42kJkg^-1K^-1#