How is the [tex]$\Delta H_{\text {fusion }}$[/tex] used to calculate the energy released when a volume of liquid freezes?
A. Liters liquid [tex]$\times 1000 ml / 1 L \times g / ml \times mol / g \times 1 / \Delta H_{\text {fusion }}$[/tex]
B. Liters liquid [tex]$\times 1000 ml / 1 L \times ml / g \times g / mol \times \Delta H_{\text {fusion }}$[/tex]
C. Liters liquid [tex]$\times 1000 ml / 1 L \times g / ml \times mol / g \times \Delta H_{\text {fusion }}$[/tex]
D. Liters liquid [tex]$\times 1000 ml / 1 L \times ml / g \times g / mol \times 1 / \Delta H_{\text {fusion }}$[/tex]

Convert liters to milliliters.
Convert milliliters to grams using density.
Convert grams to moles using molar mass.
Multiply the number of moles by Δ H fusion ​ to find the total energy released. The answer is C ​ .

Explanation

Understanding the Problem We are given four options for calculating the energy released when a volume of liquid freezes, using the enthalpy of fusion, Δ H fusion ​ . The enthalpy of fusion represents the amount of energy released or absorbed during a phase change (in this case, freezing). We need to determine which of the given formulas correctly calculates the energy released.

Steps for Calculation The correct approach involves converting the given volume of liquid (in liters) to the number of moles, and then multiplying by the enthalpy of fusion. The steps are as follows:

Convert liters (L) to milliliters (mL) using the conversion factor 1000 L mL ​ .

Convert milliliters (mL) to grams (g) using the density, which has units of mL g ​ .

Convert grams (g) to moles (mol) using the molar mass, which has units of g mol ​ .

Multiply the number of moles by Δ H fusion ​ , which has units of mol energy ​ , to find the total energy released.

Analyzing the Options Therefore, the correct formula should be:


Liters liquid × 1000 L mL ​ × mL g ​ × g mol ​ × Δ H fusion ​
Let's analyze the given options:

Option A: Liters liquid × 1000 L mL ​ × mL g ​ × g mol ​ × Δ H fusion ​ 1 ​ - Incorrect, as it divides by Δ H fusion ​ .
Option B: Liters liquid × 1000 L mL ​ × g mL ​ × mol g ​ × Δ H fusion ​ - Incorrect, as it uses g mL ​ and mol g ​ which are inverse of density and molar mass respectively.
Option C: Liters liquid × 1000 L mL ​ × mL g ​ × g mol ​ × Δ H fusion ​ - Correct, as it follows the correct conversion steps.
Option D: Liters liquid × 1000 L mL ​ × g mL ​ × mol g ​ × Δ H fusion ​ 1 ​ - Incorrect, as it uses g mL ​ , mol g ​ and divides by Δ H fusion ​ .


Conclusion Based on the analysis, the correct formula is:

Liters liquid × 1000 L mL ​ × mL g ​ × g mol ​ × Δ H fusion ​
This corresponds to option C.
Examples
Imagine you're designing a refrigeration system to freeze a certain volume of water. Knowing the enthalpy of fusion for water allows you to calculate exactly how much energy the system needs to remove to freeze that water completely. This calculation is crucial for determining the power requirements of the refrigeration unit and ensuring efficient operation. Similarly, in industrial processes involving the solidification of materials, understanding and applying the enthalpy of fusion is essential for controlling the energy balance and achieving the desired product quality.

Answered by GinnyAnswer | 2025-07-03

To compute the energy released when a liquid freezes, the process involves converting liters to milliliters, then grams, followed by moles, and finally using Δ H fusion ​ to find the energy. The correct option for this calculation is C : Liters liquid × 1000 L mL ​ × mL g ​ × g mol ​ × Δ H fusion ​ .
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Answered by Anonymous | 2025-07-04