An object with a mass of #50 g# is dropped into #160 mL# of water at #0^@C#. If the object cools by #40 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Dec 14, 2017

Answer:

The specific heat is #=2.68 kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=8ºC#

For the object #DeltaT_o=40ºC#

# 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=0.05kg#

The mass of the water is #m_w=0.16kg#

#0.05*C_o*40=0.16*4.186*8#

#C_o=(0.16*4.186*8)/(0.05*40)#

#=2.68 kJkg^-1K^-1#