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)#
