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 #16 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
May 30, 2017

Answer:

The specific heat is #=59.53kJkg^-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=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

#0.12*C_o*6=0.64*4.186*16#

#C_o=(0.64*4.186*16)/(0.12*6)#

#=59.53kJkg^-1K^-1#