An object with a mass of #120 g# is dropped into #500 mL# of water at #0^@C#. If the object cools by #30 ^@C# and the water warms by #16 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Apr 2, 2017

Answer:

The specific heat is #=9.30kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=16ºC#

For the object #DeltaT_o=30ºC#

# 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*30=0.5*4.186*16#

#C_o=(0.5*4.186*16)/(0.120*30)#

#=9.30kJkg^-1K^-1#