Calculate the heat needed to be removed to freeze 45g of water at 0c

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 "latent heat" 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, latent heat of vaporization and energy is absorbed.Vapor → liquid, latent heat of liquification and the energy is removed.Liquid → solid, latent heat of solidification and the energy is removed.Solid → liquid, latent heat of fusion 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.

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