Question #20be5

2 Answers
Mar 28, 2015

The latent heat of vaporization of ammonia is #1 * 10^(3)"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.

http://alcheme.tamu.edu/?page_id=2242

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

#q = m * DeltaH_("vap")#, where

#q# - the heat supplied;
#m# - the mass of the substance;
#DeltaH_("vap")# - the latent heat of vaporization.

Plug in your data and solve for #DeltaH_("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

#DeltaH_("vap") = color(green)(1 * 10^(3) "J/g")#

Mar 29, 2015

Answer:

#6.5/5 = 1.3 MJ kg^-1#

Explanation:

The "Latent heat of vaporization" for a substance(in this case #NH_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 MJ# caused this change at the boiling point. Which tells that ALL the heat gained = the heat needed to change in phase.
Hence, #6.5 MJ# IS the latent heat for vaporization of #NH_3#

But the question asks for "specific" latent heat,
so we divide by the mass,
# (6.5MJ)/(5kg) = 1.3 MJ kg^-1#