Chapter 20: Problem 10
A freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80 kg of water at 25.0\(^\circ\)C to 1.80 kg of ice at -5.0\(^\circ\)C in one hour. (a) What amount of heat must be removed from the water at 25.0\(^\circ\)C to convert it to ice at -5.0\(^\circ\)C? (b) How much electrical energy is consumed by the freezer during this hour? (c) How much wasted heat is delivered to the room in which the freezer sits?
Short Answer
Step by step solution
Calculate Heat to Cool Water to 0°C
Calculate Heat to Freeze Water at 0°C
Calculate Heat to Cool Ice to -5°C
Summing Total Heat Removed
Calculate Electrical Energy Consumed
Calculate Wasted Heat Delivered
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Transfer
Each stage involves different calculations and understanding of concepts. The heat transfer amount depends on both specific heat capacity and latent heat. Understanding how heat is exchanged in these processes helps us calculate the overall energy needed to transform the water.
- Cooling the Water: This phase uses the specific heat capacity of water to compute the heat extracted as water cools to 0°C.
- Freezing the Water: Latent heat is employed here, marking the energy removal needed at constant temperature to change the state from liquid to solid.
- Cooling the Ice: The specific heat of ice is used to determine heat extraction as the temperature of the newly formed ice decreases.
Coefficient of Performance
In this problem, the freezer's COP is given as 2.40. A higher COP indicates a more efficient system. According to the formula for COP, we identify how much electrical work is done compared to the heat removed.
- Formula Understanding: \[ \text{COP} = \frac{Q}{W} \]Where:
- \( Q \) is the value of total heat removed (393,858 J).
- \( W \) is the electrical energy consumed.
- Practical Implications: The calculation helps determine the work necessary to achieve the cooling process, ensuring energy effectiveness.
Latent Heat
In this exercise, latent heat plays a crucial role in transitioning water at 0°C to ice at 0°C. The latent heat of fusion for water is 334,000 J/kg.
- Role in Phase Change: This latent heat must be removed entirely for the liquid water to transform into solid ice.
- Importance: Calculating latent heat accurately ensures you understand the energy exchange in a phase change, which does not depend on the initial or final temperature, but rather the mass of the substance alone.
Specific Heat Capacity
In practical terms, it tells us how much energy is needed per kilogram to achieve a temperature increase or decrease.
- Application in this Exercise: Both water and ice have specific heat capacities that determine the energy removal amount required to cool them.
- Water has a specific heat capacity of 4186 J/kg°C.
- For ice, it is 2108 J/kg°C.
- Temperature Change Calculation: The specific heat capacity allows calculating the required energy to adjust the temperature of substances, influencing productivity and efficiency in freezing processes.