An object with a mass of #160 g# is dropped into #880 mL# of water at #0^@C#. If the object cools by #120 ^@C# and the water warms by #12 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Feb 24, 2017

Answer:

The specific heat is #=2.30kJkg^-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=12º#

For the object #DeltaT_o=120º#

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

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

#m_0 C_o*120 = m_w* 4.186 *12#

#0.160*C_o*120=0.88*4.186*12#

#C_o=(0.88*4.186*12)/(0.160*120)#

#=2.30kJkg^-1K^-1#