An object with a mass of #180 g# is dropped into #660 mL# of water at #0^@C#. If the object cools by #36 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jun 19, 2017

Answer:

The specific heat is #=1.28kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=3ºC#

For the object #DeltaT_o=36ºC#

# 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

Mass of the object is #m_0=0.18kg#

Mass of the water is #m_w=0.66kg#

#0.18*C_o*36=0.66*4.186*3#

#C_o=(0.66*4.186*3)/(0.18*36)#

#=1.28kJkg^-1K^-1#