An object with a mass of #64 g# is dropped into #270 mL# of water at #0^@C#. If the object cools by #72 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

Question

1206 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-23T08:40:12

The specific heat is #=735.8Jkg^-1K^-1#

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

Let #T=# final temperature of the object and the water

For the cold water, # Delta T_w=3º#

For the object #DeltaT_o=72º#

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

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

#m_0 C_o*70 = m_w* 4.186 *3#

#0.064*C_o*72=0.27*4.186*3#

#C_o=(0.27*4.186*3)/(0.064*72)#

#=735.8Jkg^-1K^-1#