An object with a mass of #8 g# is dropped into #20 mL# of water at #0^@C#. If the object cools by #15 ^@C# and the water warms by #60 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Apr 13, 2017

Answer:

The specific heat is #=41.86kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=60ºC#

For the object #DeltaT_o=15ºC#

# 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

#0.008*C_o*15=0.020*4.186*60#

#C_o=(0.020*4.186*60)/(0.008*15)#

#=41.86kJkg^-1K^-1#