Question

Calculate the heat energy released when 13.3 g of liquid mercury at 25.00 C is converted to solid mercury at its melting point.Constants for mercury at 1 atmheat capacity of Hg(l) 28.0 J/(mol K)melting point 234.32 Kenthalphy of fusion 2.29 KJ/mol

Answer 1

`-270.321012\ J`

Step-by-step explanation:

`C_v` = Heat capacity of Hg = 28 J/mol

`\Delta T` = Change in temperature =
`(234.32-(273.15+25))`

`\Delta H_(f)` = Enthalpy of fusion = 2.29 kJ/mol

The number of moles is given by

`n=13.3* (1)/(200.59)\\\Rightarrow n=0.0663\ molHg`

Heat is given by

`Q_1=nC_v\Delta T\\\Rightarrow Q_1=0.0663* 28* (234.32-(273.15+25))\\\Rightarrow Q_1=-118.494012\ J`

Heat released is given by

`Q_2=-n\Delta H_(f)\\\Rightarrow Q_2=-0.0663* 2.29* 10^3\\\Rightarrow Q_2=-151.827\ J`

Total heat is given by

`Q=Q_1+Q_2\\\Rightarrow Q=-118.494012+(-151.827)\\\Rightarrow Q=-270.321012\ J`

The total heat released is
`-270.321012\ J`