An object with a mass of #60 g# is dropped into #700 mL# of water at #0^@C#. If the object cools by #36 ^@C# and the water warms by #24 ^@C#, what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-06-28T09:11:31

The specific heat is #=32.56kJkg^-1K^-1#

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=36ºC#

# 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#

#0.060*C_o*36=0.7*4.186*24#

#C_o=(0.7*4.186*24)/(0.060*36)#

#=32.56kJkg^-1K^-1#