An object with a mass of #21 g# is dropped into #400 mL# of water at #0^@C#. If the object cools by #160 ^@C# and the water warms by #12 ^@C#, what is the specific heat of the material that the object is made of?

Question

1419 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-27T19:25:44

The specific heat of the object is #=5.98 kJkg^-1K^-1#

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

For the cold water, # Delta T_w=12ºC#

For the object #DeltaT_o=160ºC#

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

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

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

The mass of the object is #m_o=0.021kg#

The mass of the water is #m_w=0.4kg#

#0.021*C_o*160=0.40*4.186*12#

#C_o=(0.40*4.186*12)/(0.021*160)#

#=5.98 kJkg^-1K^-1#