An object with a mass of #10 g# is dropped into #240 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #5 ^@C#, what is the specific heat of the material that the object is made of?

Question

1307 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-13T07:38:23

The specific heat is #=4.186 kJkg^-1K^-1#

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

For the cold water, # Delta T_w=5ºC#

For the object #DeltaT_o=120ºC#

# 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=0.01kg#

The mass of the water is #m_w=0.24kg#

#0.01*C_o*120=0.24*4.186*5#

#C_o=(0.24*4.186*5)/(0.01*120)#

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