An object with a mass of #150 g# is dropped into #900 mL# of water at #0^@C#. If the object cools by #40 ^@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 14, 2018

Answer:

The specific heat is #=10.05 kJkg^-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 hot object #DeltaT_o=40ºC#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

The specific heat of water #C_w=4.186kJkg^-1K^-1#

Let, #C_o# be the specific heat of the object

The mass of the object is #m_o=0.150kg#

The mass of the water is #m_w=0.900kg#

#0.15*C_o*40=0.9*4.186*16#

#C_0=(0.9*4.186*16)/(0.15*40)#

#=10.05 kJkg^-1K^-1#