An object with a mass of #100 g# is dropped into #500 mL# of water at #0^@C#. If the object cools by #50 ^@C# and the water warms by #4 ^@C#, what is the specific heat of the material that the object is made of?

Question

1752 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-04T14:34:59

The specific heat is #=1.67kJkg^-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=4º#

For the metal #DeltaT_o=50º#

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

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

#m_0 C_o*50 = m_w* 4.186 *4#

#0.1*C_o*50=0.5*4.186*4#

#C_o=(0.5*4.186*4)/(0.1*50)#

#=1.67kJkg^-1K^-1#