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How long would a \(3.00-\mathrm{kW}\) space heater have to run to put into a kitchen the same amount of heat as a refrigerator ( (coefficient of performance \(=3.00\) ) ) does when it freezes \(1.50 \mathrm{~kg}\) of water at \(20.0^{\circ} \mathrm{C}\) into ice at \(0.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The space heater must run for approximately 69.6 seconds.

Step by step solution

01

Calculate Heat Removed by Refrigerator

To calculate the heat removed when water is converted from liquid to ice, we consider both the cooling from 20°C to 0°C, and the phase change from water to ice. First, use the formula for cooling: \[ Q = mc(T_f - T_i) \]where \( m = 1.5 \text{ kg} \) is the mass of the water, \( c = 4184 \text{ J/kg°C} \) is the specific heat of water, \( T_f = 0°C \) and \( T_i = 20°C \). Substituting the numbers:\[ Q_{cool} = 1.5 \times 4184 \times (0 - 20) = -125,520 \text{ J} \]
02

Calculate Heat of Fusion

Next, calculate the heat released during the phase change using the formula:\[ Q = mL_f \]where \( L_f = 334,000 \text{ J/kg} \) is the latent heat of fusion for water. Substituting the numbers:\[ Q_{fusion} = 1.5 \times 334,000 = 501,000 \text{ J} \]
03

Total Heat Removed

Add the heat removed during cooling and freezing:\[ Q_{total} = Q_{cool} + Q_{fusion} = |125,520| + 501,000 = 626,520 \text{ J} \]
04

Heat Input by Refrigerator

The heat input by the refrigerator, using its coefficient of performance (COP), is given by:\[ Q_{input} = \frac{Q_{total}}{\text{COP}} \]where \( \text{COP} = 3.00 \). Thus:\[ Q_{input} = \frac{626,520}{3} = 208,840 \text{ J} \]
05

Determine Time for Space Heater

We need the space heater to supply the same amount of energy to the kitchen as the refrigerator's input:\[ P = 3.00 \text{ kW} = 3000 \text{ W} \]Time \( t \) is given by \( t = \frac{Q_{input}}{P} \). So:\[ t = \frac{208,840}{3000} \approx 69.6 \text{ seconds} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Performance
The Coefficient of Performance (COP) is a measure of a refrigerator or heat pump's efficiency when transferring heat. For a refrigerator, the COP tells us how much heat it can move from the inside to the outside per unit of energy it consumes. Specifically, it is defined as the ratio of the heat removed from the cold reservoir, i.e., the inside of the fridge, to the work input. If a refrigerator has a COP of 3, this means that for every 1 Joule of electrical energy used, it can remove 3 Joules of heat. It's a way of showing that the system can transfer more energy than it consumes, making these devices energy efficient under most circumstances. The higher the COP, the more efficient the energy transfer is. When a refrigerator freezer is in use, understanding COP becomes crucial because it helps in determining both the energy usage and the time needed to perform tasks.
Specific Heat
Specific heat is a property that tells us how much energy it takes to change the temperature of 1 kg of a substance by 1°C. Essentially, it measures the amount of heat needed to increase the temperature of an object. For water, this value is particularly high at 4184 J/kg°C, meaning water requires a significant amount of energy to change temperature compared to other substances. This property of water is why it is often used as a coolant in industrial processes. In our problem, specific heat enables us to calculate how much heat energy we need to remove from water to reduce its temperature from 20°C to 0°C before it can freeze. This part of the process requires a precise understanding of specific heat, as it directly relates to the energy involved in temperature changes before any phase change occurs.
Latent Heat
Latent heat is the energy involved when a substance undergoes a phase change—such as from liquid to solid—without changing its temperature. In the context of water turning into ice, we refer to this as the latent heat of fusion. For water, this is a large value, approximately 334,000 J/kg. This high latent heat means the water will absorb a lot of energy from the surroundings before it changes phase, without any change in temperature. To freeze water, you need to remove this specific amount of latent heat per kilogram of water. In our discussed solution, it was crucial to account for latent heat to calculate the total energy the refrigerator must remove to freeze 1.5 kg of water into ice, helping us understand the overall energy transfer involved in the freezing process.
Energy Conversion
Energy conversion explains how different forms of energy—like electrical, chemical, thermal—are transformed from one type to another. In thermodynamics, we frequently encounter situations requiring such transformations to accomplish work or transfer heat. In the given exercise, the energy conversion begins with electrical energy used by the refrigerator. This energy is converted into the mechanical work needed for the refrigeration cycle, which, in turn, moves heat energy out of the water, ultimately changing from a liquid to a solid. Understanding these conversions helps to calculate how long the 3 kW space heater needs to run to output the equivalent energy as the refrigerator's input. It emphasizes the application of energy conservation principles, representing an integral part of thermodynamic calculations.

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Most popular questions from this chapter

An engine does \(18500 \mathrm{~J}\) of work and rejects \(6550 \mathrm{~J}\) of heat into a cold reservoir whose temperature is \(285 \mathrm{~K}\). What would be the smallest possible temperature of the hot reservoir?

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