An object with a mass of #18 g# is dropped into #360 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #2 ^@C#, what is the specific heat of the material that the object is made of?

Question

1267 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-18T06:11:42

The specific heat is #=1.4kJkg^-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=120ºC#

# 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

Mass of the object is #m_0=0.018kg#

Mass of the water is #m_w=0.36kg#

#0.018*C_o*120=0.36*4.186*2#

#C_o=(0.36*4.186*2)/(0.018*120)#

#=1.4kJkg^-1K^-1#