Multiply delta H (1.76 kJ/g) by mass (11.2g)
That results in 19.7 kJ, convert to J. (1 kJ=1000J)
The answer would be 19,700 J or 19.7 kJ, depending on your units.
The heat needed to melt one gram of a solid at its melting point is the heat of fusion.
Heat of vaporization
The energy needed to completely vaporize a mole of a liquid
It's not the quality of the heat, but the quantity of the heat. It means you need to put a large amount of (heat) energy into that substance to convert it from liquid to vapor at the same temperature. (Check out boiling water.)
(latent) heat of vaporization
Molar heat of fusion: the heat (enthalpy, energy) needed to transform a solid in liquid (expressed in kJ/mol). Molar heat of vaporization: the heat (enthalpy, energy) needed to transform a liquid in gas (expressed in kJ/mol).
Heat of vaporization or enthalpy of vaporization. It is the additional energy, per unit mass, required after vaporization temperature (boiling point) is reached, to accomplish the change in state, from liquid to gas.
The heat needed to melt one gram of a solid at its melting point is the heat of fusion.
thermol
heat of vaporization
It is the additional amount of heat [measured in (the "big") calories ] required to cause boiling/vaporization of a liquid already at the boiling/vaporization temperature. Heat of vaporization is the amount of heat (540 calories/gram) needed to change a specific amount of liquid (1 gram) at it's boiling point into a gas without changing it's temperature. -Qwasas
The energy needed to go from a liquid to a gas is referred to as heat of vaporization.
The latent heat of evaporation
Heat of vaporization
Heat of Vaporization id the amount of heat needed to transform a liquid into a gas while not raising its temperature.
partly boiled. The food is partly cooked and finished off when needed.
The energy needed to change a substance from a liquid to a gas is called the enthalpy (or heat) of vaporization.