An object with a mass of #64 g# is dropped into #320 mL# of water at #0^@C#. If the object cools by #16 ^@C# and the water warms by #4 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Mar 8, 2017

Answer:

The specific heat is #=5.23kJkg^-1K^-1#

Explanation:

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

For the cold water, # Delta T_w=4º#

For the object #DeltaT_o=16º#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

#C_w=4.186kJkg^-1K^-1#

#m_0 C_o*16 = m_w* 4.186 *4#

#0.064*C_o*16=0.32*4.186*4#

#C_o=(0.32*4.186*4)/(0.064*16)#

#=5.23kJkg^-1K^-1#