An object with a mass of #18 g# is dropped into #400 mL# of water at #0^@C#. If the object cools by #150 ^@C# and the water warms by #12 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Dec 28, 2017

Answer:

#=(7.4311J)/(g*C)#

Explanation:

  1. Given the volume of water used as medium to cool down the object, its mass can be derived through the density formula; that is,
    #rho=m/V#
    #m=rhoxxV#
    #m=(1g)/cancel((ml))xx400cancel(ml)#
    #m=400gH_2O#
  2. Find the heat absorbed (gained) by the water that increases its temperature by #12^o#. Knowing the #Cp_w=(4.18J)/(g*C)#, heat gained can be calculated by:
    #Q=mCpDeltaT#
    #Q=(400cancel(g))((4.18J)/cancel((g*C)))(12)cancel(^oC)#
    #Q=20,064J=20.064kJ#
  3. Since heat lost=heat gained; therefore, the #Cp# of the material can be computed as;
    #Q=mCpDeltaT#
    #Cp_(o)=(Q)/(mDeltaT)#
    #Cp_(o)=(20,064J)/((18g)(150^oC))#
    #Cp_(o)=(7.4311J)/(g*C)#