An object with a mass of #24 g# is dropped into #6400 mL# of water at #0^@C#. If the object cools by #60 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jan 25, 2018

Answer:

The specific heat is #=55.8 kJkg^-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=60ºC#

# 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=0.024kg#

The mass of the water is #m_w=6.4kg#

#0.024*C_o*60=6.4*4.186*3#

#C_o=(6.4*4.186*3)/(0.024*60)#

#=55.8 kJkg^-1K^-1#