An object with a mass of #180 g# is dropped into #660 mL# of water at #0^@C#. If the object cools by #36 ^@C# and the water warms by #9 ^@C#, what is the specific heat of the material that the object is made of?

Question

1681 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-05T14:45:32

The specific heat is #=3.84kJkg^-1K^-1#

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

For the cold water, # Delta T_w=9º#

For the metal #DeltaT_o=36º#

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

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

#m_0 C_o*36 = m_w* 4.186 *9#

#0.18*C_o*36=0.66*4.186*9#

#C_o=(0.66*4.186*9)/(0.18*36)#

#=3.84kJkg^-1K^-1#