An object with a mass of #90 g# is dropped into #500 mL# of water at #0^@C#. If the object cools by #30 ^@C# and the water warms by #30 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jul 19, 2017

Answer:

The specific heat is #=23.3kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=30^@C#

For the object #DeltaT_o=30^@C#

# 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=0.090kg#

Mass of the water is #m_w=0.5kg#

#0.09*C_o*30=0.5*4.186*30#

#C_o=(0.5*4.186*30)/(0.09*30)#

#=23.3kJkg^-1K^-1#