An object with a mass of #150 g# is dropped into #850 mL# of water at #0^@C#. If the object cools by #40 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Mar 3, 2017

Answer:

The specific heat of the object is #=1.78kJkg^-1K^-1#

Explanation:

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

For the cold water, # Delta T_w=3º#

For the metal #DeltaT_o=40º#

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

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

#m_0 C_o*40 = m_w* 4.186 *3#

#0.15*C_o*40=0.85*4.186*3#

#C_o=(0.85*4.186*3)/(0.15*40)#

#=1.78kJkg^-1K^-1#