Answer to Question #91772 in General Chemistry for Desma Palacios

Question

Calculate the heat energy released when 13.0 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point.
Constants for mercury at 1 atm:
heat capacity of Hg(l); 28.0 J/(mol⋅K)
melting point; 234.32 K
enthalpy of fusion; 2.29 kJ/mol

Answer 1

Liquid mercury will release some of its internal energy to the surroundings when it cools and then solidifies.

Cooling: Q = mcΔT

T = 25 + 273 = 298 K

c = 28.0 J/(mol K) / 201 g/mol = 0.14 J/(g K)

L = 2.29 kJ/mol / 201 g/mol = 0.0114 kJ/g

Q = mcΔT = (13.0 g)(0.14 J/(g K))(234.32 K – 298 K) = -116 J

Solidification: Q = mL = (13.0 g)(-0.0114 kJ/g) = -0.1482 kJ = -148 J

The total energy released is Q = (-116 J) + (-335 J) = -451 J

The negative sign represents energy lost by the system.