An object with a mass of #30 g# is dropped into #750 mL# of water. If the object cools by #15 ^@C# and the water warms by #15 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Mar 1, 2017

Answer:

The specific heat is #=104.65kJkg^-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=15º#

For the object #DeltaT_o=15º#

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

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

#m_0 C_o*15 = m_w* 4.186 *15#

#0.030*C_o*15=0.75*4.186*15#

#C_o=(0.75*4.186*15)/(0.030*15)#

#=104.65kJkg^-1K^-1#