# 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

#### Explanation:

- 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# - 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# - 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)#