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Alenkasestr [34]
3 years ago
10

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.C

onstants 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
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

-270.321012\ J

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\times \dfrac{1}{200.59}\\\Rightarrow n=0.0663\ molHg

Heat is given by

Q_1=nC_v\Delta T\\\Rightarrow Q_1=0.0663\times 28\times (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\times 2.29\times 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

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