An object with a mass of #32 g# is dropped into #480 mL# of water at #0^@C#. If the object cools by #80 ^@C# and the water warms by #6 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Feb 27, 2017

Answer:

The specific heat is #=4.71kJkg^-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=6º#

For the object #DeltaT_o=80º#

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

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

#m_0 C_o*80 = m_w* 4.186 *6#

#0.032*C_o*80=0.48*4.186*6#

#C_o=(0.48*4.186*6)/(0.032*80)#

#=4.71kJkg^-1K^-1#