# Calculate the heat energy released when 18.1g of liquid mercury at 25.00 degrees C is converted...

## Question:

Calculate the heat energy released when 18.1g of liquid mercury at 25.00 degrees C is converted to solid mercury at its melting point. Constants for mercury at 1 atm:

Heat capacity of Hg(l): 28.0J/(mol*K)

Melting point 234.32 K

Enthalpy of fusion 2.29 kJ/mol.

## Phase Transition:

During a phase transition, the temperature is not altered. The heat transferred are involved in phase transition and can be calculated using the enthalpy of phase transition. For exothermic processes such as freezing and condensation, the enthalpy involved would have a negative sign.

## Answer and Explanation:

**The total heat released is 0.368 kJ.**

First let's translate the initial temperature of Hg(l) to units of Kelvin

$$25.00 ^oC + 273.15 = 298.15 \ K $$

The we need to convert the heat capacity to {eq}\dfrac {J}{g \cdot K} {/eq}

$$28.0 \dfrac {J}{mole \cdot K} \times \dfrac {1 \ mole}{200.59 \ g} = 0.1396 \dfrac {J}{g \cdot K} $$

The initial step is to cool the Hg from 298.15 K to it's freezing point of 234.32 K. The heat transferred (q) involved is calculated using the specific heat (C) of liquid Hg

$$q = mc \Delta T\\ q = (18.1 \ g)(0.1396 \dfrac {J}{g \cdot K})(234.32 - 298.15 \ K)\\ q = -161.28 \ J = -0.16128 \ kJ \ $$

Once at the freezing point, the energy released will be the result of the freezing process at a constant temperature. The calculation would involve the enthalpy of fusion, but with a negative sign.

$$18.1 \ g \ Hg \times \dfrac {1 \ mole \ Hg}{200.59 \ g \ Hg} \times \dfrac {-2.29 \ kJ}{mole} = -0.20664 \ kJ $$

Now adding the heat transferred of both process together will give us the total heat energy released.

$$-0.16128 \ kJ + -0.20664 \ kJ = \boxed {-0.368 \ kJ} $$

The total heat released is 0.368 kJ.

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from College Chemistry: Help and Review

Chapter 6 / Lesson 5