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.
Answer:
I'm pretty sure it's neutrons the first one
