An object with a mass of #60 g# is dropped into #800 mL# of water at #0^@C#. If the object cools by #30 ^@C# and the water warms by #15 ^@C#, what is the specific heat of the material that the object is made of?

Question

1775 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-18T14:28:25

The specific heat is #=27.91kJkg^-1K^-1#

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

For the cold water, # Delta T_w=15^@C#

For the object #DeltaT_o=30^@C#

# 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=0.060kg#

Mass of the water is #m_w=0.8kg#

#0.06*C_o*30=0.8*4.186*15#

#C_o=(0.8*4.186*15)/(0.06*30)#

#=27.91kJkg^-1K^-1#