Answer: The heat needed to be removed to freeze 45.0 g of water at 0.0 °C is 15.01 KJ.
Explanation:
Firstly, we need to define the term <em>"latent heat"</em> which is the amount of energy required "absorbed or removed" to change the phase "physical state; solid, liquid and vapor" without changing the temperature.
Types of latent heat: depends on the phases that the change occur between them;
Liquid → vapor, <em>latent heat of vaporization</em> and energy is absorbed.
Vapor → liquid, latent heat of liquification and the energy is removed.
Liquid → solid, <em>latent heat of solidification</em> and the energy is removed.
Solid → liquid, <em>latent heat of fusion</em> and the energy is absorbed.
In our problem, we deals with latent heat of freezing "solidification" of water.
The latent heat of freezing of water, ΔHf, = 333.55 J/g; which means that the energy required to be removed to convert 1.0 g of water from liquid to solid "freezing" is 333.55 g at 0.0 °C.
Then the amount of energy needed to be removed to freeze 45.0 g of water at 0.0 °C is (ΔHf x no. of grams of water) = (333.55 J/g)(45.0 g) = 15009.75 J = 15.01 KJ.
Steam distillation is a separation process for temperature sensitive materials and it is used for purification of organic compounds.
The water vapor carries small amounts of the vaporized compounds (water and organic molecules) to the condensation flask. This is preferred technique because allows easy separation of the components by decantation or other suitable methods, because product od distilation is a two phase system<span> of water and the organic distillate.</span>
