An object with a mass of #8 g# is dropped into #50 mL# of water at #0^@C#. If the object cools by #5 ^@C# and the water warms by #70 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Dec 8, 2017

Answer:

Thespecific heat is #=366.3kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=70ºC#

For the object #DeltaT_o=5ºC#

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

tTe 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=0.008kg#

The mass of the water is #m_w=0.050kg#

#0.008*C_o*5=0.050*4.186*70#

#C_o=(0.050*4.186*70)/(0.008*5)#

#=366.3 kJkg^-1K^-1#