An object with a mass of #120 g# is dropped into #640 mL# of water at #0^@C#. If the object cools by #20 ^@C# and the water warms by #16 ^@C#, what is the specific heat of the material that the object is made of?

Question

1272 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-20T10:30:53

The specific heat is #=17.86kJkg^-1K^-1#

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

For the cold water, # Delta T_w=16^@C#

For the object #DeltaT_o=20^@C#

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

The specific heat of water is #C_w=4.186kJkg^-1K^-1#

Let, #C_o# be the specific heat of the object

Mass of the object is #m_0=0.120kg#

Mass of the water is #m_w=0.64kg#

#0.12*C_o*20=0.64*4.186*16#

#C_o=(0.64*4.186*16)/(0.12*20)#

#=17.86kJkg^-1K^-1#