An object with a mass of #64 g# is dropped into #210 mL# of water at #0^@C#. If the object cools by #96 ^@C# and the water warms by #7 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jul 14, 2017

Answer:

#1.00*10^(-3)kJkg^(-1)K^(-1)#

Explanation:

Object:
Mass - #64g=0.064kg#
Specific heat cpacity - ?
Temperature change = #96K#

Water:
Mass - #210mL=0.00021m^3~~0.00021kg#.
Specific heat capacity - #4.18 kJ K^(-1) kg^(-1)#
Temperature change = #7K#

#m_oC_oDeltaT_o=m_wC_wDeltaT_w#

#(0.064*96)C_o=(4.18*7*0.00021)#

#C_o=(4.18*7*0.00021)/(0.064*96)=0.00100009766~~1.00*10^(-3)kJkg^(-1)K^(-1)#