An object with a mass of #60 g# is dropped into #700 mL# of water at #0^@C#. If the object cools by #40 ^@C# and the water warms by #24 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jul 16, 2017

Answer:

The specific heat is #=29.3kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=24ºC#

For the object #DeltaT_o=40ºC#

# 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=0.060kg#

Mass of the water is #m_w=0.7kg#

#0.06*C_o*40=0.7*4.186*24#

#C_o=(0.7*4.186*24)/(0.06*40)#

#=29.3kJkg^-1K^-1#