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10.3.5. Example of Latent Heat Calculation
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Today we're going to dive into latent heat, specifically the latent heat of fusion. This is the heat required to change a solid into a liquid without changing its temperature. Can anyone give an example of a substance that undergoes this change?
Ice melting into water!
Exactly! So, when ice melts at 0°C, it absorbs heat without rising in temperature. This heat is called the latent heat of fusion, and for ice, it's 334 kJ/kg. Remember this number; a nice acronym is 'FUSE'—F for Fusion, U for Up (as in temperature remains the same), S for Solid to liquid, and E for Energy needed.
Got it! So, if I have 1 kg of ice, I would need 334 kJ of heat to melt it, right?
Excellent! So how much heat would you need for 0.5 kg of ice?
That would be 167 kJ.
Correct! 0.5 kg times 334 kJ/kg gives us 167 kJ. Always remember: for half the mass, halve the energy!
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Next, let’s look at the latent heat of vaporization. This is the heat required to turn a liquid into a gas without changing the temperature. Anyone know the latent heat of vaporization for water?
Isn't it 2260 kJ/kg?
Spot on! Now, using the formula Q = mL, if you wanted to vaporize 0.2 kg of water, how much heat would you need?
I would calculate Q as 0.2 kg multiplied by 2260 kJ/kg, which equals 452 kJ.
Right again! That means you need 452 kJ to turn 0.2 kg of water at 100°C into steam—keep those numbers handy!
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Create a free accountLet's apply what we've learned. Imagine you're making ice cream using an ice mixture. How does the principle of latent heat come into play?
The ice absorbs heat as it melts, which cools the mixture down without raising its temperature!
Exactly! The melting ice helps keep the ice cream base cool while absorbing heat, thus utilizing the latent heat of fusion beautifully in the process.
So it’s about the balance of energy transfer, not just the temperature!
Very insightful! We'll see this energy balance in other scenarios, such as boiling water, where heat absorbed leads to a phase change.
Overview
Short Summary
This section explains the calculation of latent heat, illustrating how it is used in phase changes without a temperature change.
Medium Summary
The section provides a detailed analysis of latent heat calculation, focusing on the latent heat of fusion and vaporization through examples. It highlights how to apply the formula Q=mL to find the heat required for phase changes.
Detailed Summary
Example of Latent Heat Calculation
This section covers the essential concept of latent heat and its calculation in phase changes. Latent heat is defined as the heat needed to change the state of a substance without altering its temperature. Two critical types are elaborated:
Types of Latent Heat
- Latent Heat of Fusion: Heat required for a solid to become liquid at a constant temperature (e.g., melting of ice).
- Latent Heat of Vaporization: Heatrequired for a liquid to become a gas at constant temperature (e.g., boiling of water).
Formula
The general formula used for calculating latent heat is:
Q = mL
- Q = Heat energy (in Joules)
- m = Mass of the substance (in kilograms)
- L = Latent heat of fusion or vaporization (in J/kg)
Examples
An example provided is the calculation of heat required to melt 500 grams of ice at 0°C using the latent heat of fusion. This calculation facilitates understanding of how to apply the formula in practical scenarios. The latent heat of fusion of ice is specified as 334 kJ/kg, leading to:
Q = 0.5 × 334 = 167 kJ
This shows that 167 kJ of heat is necessary to melt 500 grams of ice.
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Create a free accountTo melt 500 grams of ice at 0°C, using the latent heat of fusion: Q=0.5×334=167 kJ Q = 0.5 imes 334 = 167 , \text{kJ}
Detailed Explanation
In this example, we are calculating how much heat energy is needed to melt ice. The formula used is Q = mL, where Q is the heat energy required, m is the mass of the substance (in kilograms), and L is the latent heat of fusion (in J/kg). In this case, 500 grams of ice needs to be melted, which is equal to 0.5 kilograms. The latent heat of fusion for ice is given as 334 kJ/kg. Therefore, by multiplying the mass (0.5 kg) by the latent heat (334 kJ/kg), we find that 167 kJ of heat is required to melt the ice.
Examples & Analogies
Imagine you are trying to make ice cubes melt to fill a glass with cold water. You notice that when you put ice in the glass, it doesn't immediately turn into water but instead just sits there until it has absorbed enough heat. This is because the ice requires a specific amount of heat to change its state from solid to liquid without a temperature change. In total, to melt the 500 grams of ice, you would need to use 167 kJ of energy from your surroundings to successfully complete this state change.
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Key Concepts
Core takeaways and short definitions to help you quickly recall the key ideas from this section.
Latent Heat: Heat absorbed during a phase change without temperature change.
Latent Heat of Fusion: Specifically pertains to the solid-to-liquid transition.
Latent Heat of Vaporization: Pertains to the liquid-to-gas transition.
Examples
Step-by-step examples to apply the section's ideas and test your understanding.
An example provided is the calculation of heat required to melt 500 grams of ice at 0°C using the latent heat of fusion. This calculation facilitates understanding of how to apply the formula in practical scenarios. The latent heat of fusion of ice is specified as 334 kJ/kg, leading to:
Q = 0.5 × 334 = 167 kJ
This shows that 167 kJ of heat is necessary to melt 500 grams of ice.
Memory Aids
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Glossary
Latent Heat
Heat required to change the state of a substance without changing its temperature.
Latent Heat of Fusion
The heat required to change a substance from solid to liquid at constant temperature.
Latent Heat of Vaporization
The heat required to change a substance from liquid to gas at constant temperature.