# Question 20be5

Mar 28, 2015

The latent heat of vaporization of ammonia is $1 \cdot {10}^{3} \text{J/g}$.

The latent heat of vaporization represents the energy you must supply, in the form of heat, to a substance to change it from liquid to vapor.

You're basically dealing with a phase change, during which you supply heat to the liquid ammonia to transform it to steam - at constant temperature.

Mathematically, the amount of heat required to vaporize a certain mass of a substance is given by

$q = m \cdot \Delta {H}_{\text{vap}}$, where

$q$ - the heat supplied;
$m$ - the mass of the substance;
$\Delta {H}_{\text{vap}}$ - the latent heat of vaporization.

Plug in your data and solve for $\Delta {H}_{\text{vap}}$ by

DeltaH_("vap") = q/m = (6.5 * 10^(6)"J")/(5 * 10^(3)"g") = 1.3 * 10^(3)"J/g"#

If you round this to one sig fig, the number of sig figs given for 5 kg, the answer will be

$\Delta {H}_{\text{vap") = color(green)(1 * 10^(3) "J/g}}$

Mar 29, 2015

$\frac{6.5}{5} = 1.3 M J k {g}^{-} 1$

#### Explanation:

The "Latent heat of vaporization" for a substance(in this case $N {H}_{3}$) is the heat that is required to cause a change in phase of a given mass(not specified)

We are told that $6.5 M J$ caused this change at the boiling point. Which tells that ALL the heat gained = the heat needed to change in phase.
Hence, $6.5 M J$ IS the latent heat for vaporization of $N {H}_{3}$

But the question asks for "specific" latent heat,
so we divide by the mass,
$\frac{6.5 M J}{5 k g} = 1.3 M J k {g}^{-} 1$