calculate the heat energy released

Question

Calculate the heat energy released when 13.7 C 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 x K)

melting point - 234.32 K

enthalpy of fusion - 2.29 kJ/mol

Answer 1

I'm assuming there is a typo here, where 13.7 C should read 13.7 grams.

Two things are happening here: (1) we are lowering the temperature of the mercury, and (2) we are changing the phase of the mercury from liquid to solid. Importantly, these two thing occur in two different steps, and we can calculate the heat released from both of them using formulas for heat capacity, and enthalpy of fusion.

(1) The melting point of mercury is 234.32 K. We start at 25.00°C, which is 298.15 K. Since we are lowering the temperature of a pure phase, we can use the formula

q = mcΔT

m = 13.7 g

c = 28.0 J / mol•K

ΔT = (234.32 K – 298.15 K)

Convert grams to moles:

13.7 g Hg * (1 mol Hg / 200.59 g) = 0.0683 mol Hg

And plug in:

q = (0.0683 mol)(28.0 J / mol•K)(–63.83 K) = –122 J

(2) The enthalpy of fusion (going from solid to liquid) is 2.29 kJ / mol, so going from liquid to solid will be -2.29 kJ/mol. Since we have 0.0683 mol Hg, this means:

q = 0.0683 mol * (-2.29 kJ / 1 mol) = -0.156 kJ = –156 J

Adding (1) and (2) together, we get

-122 + -156 = -278 J

So 278 J of energy were released.