A freezer has a coefficient of performance of 6.30. The freezer is advertised as using 495 kW-h/y. Note: One kilowatt-hour (kW-h) is an amount of energy equal to operating a 1-kW appliance for one hour. (a) On average, how much energy does the freezer use in a single day? 1.356 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. J (b) On average, how much thermal energy is removed from the freezer each day? 0.215 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. J (c) What maximum amount of water at 21.0°C could the freezer freeze in a single day? (The latent heat of fusion of water is 3.33 x 105 J/kg, and the specific heat of water is 4186 J/kg. K.) kg

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### Understanding a Freezer's Energy Usage

A freezer has a coefficient of performance (COP) of 6.30. It is advertised as using 495 kW-h annually. Note that one kilowatt-hour (kW-h) is the amount of energy equal to operating a 1-kW appliance for one hour.

#### Calculations and Questions:

**(a) On average, how much energy does the freezer use in a single day?**

![Incorrect Answer]
*Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.*

**Explanation:**
To find the daily energy usage:
- Annual energy consumption = 495 kWh/year
- Number of days in a year = 365 days
\[ \text{Daily energy consumption} = \frac{495 \text{ kWh}}{365 \text{ days}} \]

**(b) On average, how much thermal energy is removed from the freezer each day?**

![Incorrect Answer]
*Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.*

**Explanation:**
Given the coefficient of performance (COP):
\[ \text{COP} = \frac{\text{Thermal energy removed}}{\text{Electrical energy input}} \]

\[ \text{Thermal energy removed} = \text{COP} \times \text{Daily energy consumption} \]

**(c) What maximum amount of water at 21.0°C could the freezer freeze in a single day?**
- **Given Data:**
  - Latent heat of fusion of water: \(3.33 \times 10^5 \) J/kg
  - Specific heat of water: 4186 J/(kg·K)

\[ \text{Thermal energy required: Q} = \text{Latent heat of fusion} \times \text{mass of water} + \text{specific heat} \times \text{mass of water} \times \text{temperature change} \]

**Explanation:**
Use the thermal energy removed calculated in part (b) to determine the mass of water that can be frozen.

### Conclusion
The above calculations involve step-by-step understanding of energy usage and thermal dynamics, reflecting practical application of physical principles for a freezer’s operation. If basic concepts and problem solving are applied correctly, accurate results are attainable.
Transcribed Image Text:### Understanding a Freezer's Energy Usage A freezer has a coefficient of performance (COP) of 6.30. It is advertised as using 495 kW-h annually. Note that one kilowatt-hour (kW-h) is the amount of energy equal to operating a 1-kW appliance for one hour. #### Calculations and Questions: **(a) On average, how much energy does the freezer use in a single day?** ![Incorrect Answer] *Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.* **Explanation:** To find the daily energy usage: - Annual energy consumption = 495 kWh/year - Number of days in a year = 365 days \[ \text{Daily energy consumption} = \frac{495 \text{ kWh}}{365 \text{ days}} \] **(b) On average, how much thermal energy is removed from the freezer each day?** ![Incorrect Answer] *Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.* **Explanation:** Given the coefficient of performance (COP): \[ \text{COP} = \frac{\text{Thermal energy removed}}{\text{Electrical energy input}} \] \[ \text{Thermal energy removed} = \text{COP} \times \text{Daily energy consumption} \] **(c) What maximum amount of water at 21.0°C could the freezer freeze in a single day?** - **Given Data:** - Latent heat of fusion of water: \(3.33 \times 10^5 \) J/kg - Specific heat of water: 4186 J/(kg·K) \[ \text{Thermal energy required: Q} = \text{Latent heat of fusion} \times \text{mass of water} + \text{specific heat} \times \text{mass of water} \times \text{temperature change} \] **Explanation:** Use the thermal energy removed calculated in part (b) to determine the mass of water that can be frozen. ### Conclusion The above calculations involve step-by-step understanding of energy usage and thermal dynamics, reflecting practical application of physical principles for a freezer’s operation. If basic concepts and problem solving are applied correctly, accurate results are attainable.
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