An object with a mass of #18 g# is dropped into #360 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #6 ^@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 #=4.186kJkg^-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=120ºC#

# 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

#0.018*C_o*90=0.36*4.186*6#

#C_o=(0.36*4.186*6)/(0.018*120)#

#=4.186kJkg^-1K^-1#