The heat of vaporization of water at 100^∘ C is 2.26 kJ / g at 37^∘ C (body temperature), it is

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The heat of vaporization of water at 100^∘ C is 2.26 kJ / g...

The heat of vaporization of water at $100^{\circ} \mathrm{C}$ is $2.26 \mathrm{~kJ} / \mathrm{g}$ at $37^{\circ} \mathrm{C}$ (body temperature), it is $2.41 \mathrm{~kJ} / \mathrm{g}$. (a) Convert the latter value to standard molar heat of vaporization, $\triangle H_{\text {vap }}^{\circ},$ at $37^{\circ} \mathrm{C}$ (b) Why is the heat of vaporization greater at $37^{\circ} \mathrm{C}$ than at $100^{\circ} \mathrm{C} ?$

Key Concepts

Heat of Vaporization

This concept refers to the amount of energy required to transform a given mass of a liquid into vapor at a constant temperature and pressure. It reflects the energy needed to overcome intermolecular forces within the liquid. In physical chemistry, it is often expressed per unit mass or per mole, highlighting its dependence on both the substance's properties and the conditions under which the phase transition occurs.

Molar Conversion

Molar conversion is the process of translating a quantity given per unit mass into a per-mole basis using the molar mass of a substance. In thermodynamic calculations, this conversion is essential because enthalpies and other energy measures are typically standardized per mole, which provides a consistent basis for comparing energy changes across different substances and conditions.

Temperature Dependence of Enthalpy of Vaporization

The enthalpy of vaporization varies with temperature because the strength of intermolecular forces and the required energy to overcome them change as temperature changes. At lower temperatures, molecules have less kinetic energy, and stronger intermolecular attractions must be overcome, resulting in a higher heat of vaporization per gram or per mole. Conversely, at higher temperatures, the molecules are already more energetic, so less additional energy is needed to vaporize the liquid.

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