An object with a mass of #64 g# is dropped into #270 mL# of water at #0^@C#. If the object cools by #96 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Apr 19, 2017

Answer:

The specific heat is #=0.55kJkg^-1K^-1#

Explanation:

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

For the cold water, # Delta T_w=3ºC#

For the object #DeltaT_o=96ºC#

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

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

Let, #C_o# be the specific heat of the object

#0.064*C_o*96=0.27*4.186*3#

#C_o=(0.27*4.186*3)/(0.064*96)#

#=0.55kJkg^-1K^-1#