What is the heat of fusion of mercury in J/kg?

When 1.00 kg of solid mercury at it’s melting point of -39.0°C is placed in an calorimeter filled with 1.20 kg of water at 20.0°C, the final temperature of the combination is found to be 16.5°C. What is the heat of fusion of mercury in J/kg? The specific heat of liquid mercury is 140.0 J/kg°C.

1 Answer
May 26, 2018

Let #L_m# be heat of fusion of mercury.
In the question #1.00\ kg# of solid mercury at #-39^@C# becomes liquid mercury by change of state at #-39^@C#. Heat required for this

#Q_1=1.00L_m\ J#

In the second step the temperature of mercury changes from #-39^@C# to #16.5^@C#. Heat required for this

#Q_2=msDeltaT#
#=>Q_2=1.00xx140.0xx(16.5-(-39))#
#=>Q_2=140.0xx55.5#
#=>Q_2=7770\ J#

Total heat gained by mercury #=Q_1+Q_2=(7770+L_m)\ J#

Assuming that there is no loss or gain of heat by calorimeter this heat is supplied by water. Using Law of conservation of energy we get

#(7770+L_m)=1.20xx4186 xx(20-16.5)#
where #4186 Jcdotkg^-1"^@C^-1# is specific heat of water
#=>(7770+L_m)=17581.2#
#=>L_m=9811.2\ Jcdot kg^-1#

#"^$#Compares well with value #=11.3\ kJcdotkg^-1#