An object with a mass of #4 g# is dropped into #650 mL# of water at #0^@C#. If the object cools by #300 ^@C# and the water warms by #2 ^@C#, what is the specific heat of the material that the object is made of?

Question

1290 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-20T12:33:13

The specific heat is #=4.53kJkg^-1K^-1#

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

For the cold water, # Delta T_w=2ºC#

For the object #DeltaT_o=300ºC#

# 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

The mass of the object is #m=0.004kg#

#0.004*C_o*300=0.650*4.186*2#

#C_o=(0.650*4.186*2)/(0.004*300)#

#=4.53kJkg^-1K^-1#