An object with a mass of #64 g# is dropped into #320 mL# of water at #0^@C#. If the object cools by #16 ^@C# and the water warms by #6 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Mar 24, 2017

Answer:

The specific heat is #=7.85kJkg^-1k^-1#

Explanation:

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

For the cold water, # Delta T_w=6ºC#

For the object #DeltaT_o=16ºC#

# 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

#0.064*C_o*16=0.32*4.186*6#

#C_o=(0.32*4.186*6)/(0.064*16)#

#=7.85kJkg^-1k^-1#