An object with a mass of #20 g# is dropped into #150 mL# of water at #0^@C#. If the object cools by #90 ^@C# and the water warms by #15 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
May 30, 2017

Answer:

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

Explanation:

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

For the cold water, # Delta T_w=15ºC#

For the object #DeltaT_o=90ºC#

# 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

#0.020*C_o*90=0.15*4.186*15#

#C_o=(0.15*4.186*15)/(0.020*90)#

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