An object with a mass of #32 g# is dropped into #560 mL# of water at #0^@C#. If the object cools by #80 ^@C# and the water warms by #6 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Dec 5, 2017

Answer:

The specific heat is #=5.49 kJkg^-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=80ºC#

# 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=0.032kg#

The mass of the water is #m_w=0.56kg#

#0.032*C_o*80=0.56*4.186*6#

#C_o=(0.56*4.186*6)/(0.032*80)#

#=5.49 kJkg^-1K^-1#