To calculate the heat required to melt a specific amount of ice, you use a simple formula that involves the mass of the ice and the heat of fusion of ice.
The heat required (Q) is calculated by multiplying the mass of the ice (m) by the heat of fusion of ice (Lf).
Understanding the Formula: Q = m × Lf
This formula quantifies the energy needed to change a substance from a solid to a liquid state at its melting point without changing its temperature.
- Q: Represents the heat energy required (often measured in Joules).
- m: Represents the mass of the substance being melted (e.g., in grams).
- Lf: Represents the heat of fusion of the substance (energy per unit mass, often measured in Joules per gram or Joules per kilogram).
What is the Heat of Fusion of Ice?
The heat of fusion (Lf) is a specific property of a substance. For ice, it's the amount of energy needed to melt one unit of mass of ice at its melting point (0°C or 32°F) into water at the same temperature.
- The commonly accepted value for the heat of fusion of ice is approximately 334 Joules per gram (J/g) or 334,000 Joules per kilogram (J/kg).
- This value is determined experimentally through calorimetry.
Performing the Calculation
Using the formula Q = m × Lf, you can easily find the heat needed.
- Identify the Mass (m): Determine the amount of ice you need to melt in grams or kilograms.
- Use the Heat of Fusion (Lf): Use the known value for the heat of fusion of ice (e.g., 334 J/g).
- Multiply: Calculate Q by multiplying the mass by the heat of fusion value.
Example Calculation
Let's use an example calculation similar to the one shown in the provided reference video (0:14 - 1:43):
Suppose you have 10 grams of ice that you want to melt.
- Mass of ice (m) = 10 grams
- Heat of fusion of ice (Lf) ≈ 334 Joules/gram
Using the formula Q = m × Lf:
Q = 10 g × 334 J/g
Q = 3340 Joules
Therefore, 3340 Joules of heat energy are required to melt 10 grams of ice at its melting point. As demonstrated in the reference, you simply "plug in 10 G. And multiply it by 334 Jew per gam."
Here's a summary table for clarity:
| Variable | Description | Common Units | Value in Example |
|---|---|---|---|
| Q | Heat Energy Required | Joules (J) | 3340 J |
| m | Mass of Ice | Grams (g) | 10 g |
| Lf | Heat of Fusion of Ice (Constant value) | J/g | 334 J/g |
This method allows you to calculate the specific amount of heat energy needed to transform a given mass of ice into liquid water without a temperature change.
