/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 A freezer has a coefficient of p... [FREE SOLUTION] | 91Ó°ÊÓ

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A freezer has a coefficient of performance of \(2.40 .\) The freezer is to convert 1.80 \(\mathrm{kg}\) of water at \(25.0^{\circ} \mathrm{C}\) to 1.80 \(\mathrm{kg}\) of ice at \(-5.0^{\circ} \mathrm{C}\) in one hour. (a) What amount of heat must be removed from the water at \(25.0^{\circ} \mathrm{C}\) to convert it to ice at \(-5.0^{\circ} \mathrm{C} ?\) (b) How much much wasted heat is delivered to the room in which the freezer sits?

Short Answer

Expert verified
(a) Remove approximately 741.9 kJ of heat. (b) Deliver around 1043 kJ of heat to the room.

Step by step solution

01

Identify Heat Required for Cooling

To find the total amount of heat that must be removed, we need to consider the cooling and freezing processes. First, cool the water from 25°C to 0°C. The heat removed in this step is calculated using: \[ Q_1 = m c_w \Delta T_1 \]where \( m = 1.80 \text{ kg} \), \( c_w = 4.186 \text{ kJ/kg°C} \) (specific heat capacity of water), and \( \Delta T_1 = 25 - 0 = 25°C \). Substitute the values to get:\[ Q_1 = 1.80 \times 4.186 \times 25 \text{ kJ} \].
02

Calculate Heat of Fusion

Next, calculate the heat required to freeze the water at 0°C into ice. The heat removed in this step is given by:\[ Q_2 = m L_f \]where \( L_f = 334 \text{ kJ/kg} \) (latent heat of fusion for water). Substitute the mass to get:\[ Q_2 = 1.80 \times 334 \text{ kJ} \].
03

Cool Ice to -5°C

Finally, cool the ice from 0°C to -5°C. The heat removed in this step is:\[ Q_3 = m c_i \Delta T_2 \]where \( c_i = 2.09 \text{ kJ/kg°C} \) (specific heat capacity of ice), and \( \Delta T_2 = 0 - (-5) = 5°C \). Substitute the values to get:\[ Q_3 = 1.80 \times 2.09 \times 5 \text{ kJ} \].
04

Calculate Total Heat Removed

The total amount of heat removed is the sum of the three individual heats calculated:\[ Q_{total} = Q_1 + Q_2 + Q_3 \].Substitute the values from the previous steps:\[ Q_{total} = (1.80 \times 4.186 \times 25) + (1.80 \times 334) + (1.80 \times 2.09 \times 5) \text{ kJ} \].Calculate \( Q_{total} \).
05

Calculate Work Done by Freezer

Using the coefficient of performance (COP) of the freezer, we can calculate the work input required to remove this amount of heat:\[ \text{COP} = \frac{Q_{cold}}{W} \Rightarrow W = \frac{Q_{total}}{\text{COP}} \].Substitute \( Q_{total} \) and COP = 2.40 to calculate work \( W \).
06

Calculate Wasted Heat Delivered to Room

The heat released to the room is the sum of the heat removed and the work input:\[ Q_{room} = Q_{total} + W \].Use the previously calculated values to find \( Q_{room} \).

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics and plays a crucial role in the operation of devices such as freezers. When a freezer removes heat, it transfers energy from the water inside, transforming it from one state to another. This process involves several steps, each requiring the removal of a specific amount of heat.

- **Cooling Water:** First, we must cool the water from its initial temperature of 25°C down to 0°C. This is achieved by removing thermal energy, which is calculated using the formula for heat transfer: \( Q = mc_w\Delta T \) where \( m \) is the mass of the water, \( c_w \) is the specific heat capacity of water, and \( \Delta T \) is the change in temperature. - **Freezing Water:** Next, to transform the water into ice at 0°C, the freezer extracts the latent heat of fusion. This is the energy required to change the state from liquid to solid without changing temperature.- **Cooling Ice:** Finally, the ice is cooled further to -5°C. This requires removing additional heat calculated using the specific heat capacity of ice. Each phase of the process requires careful consideration of the specific heat and latent heat involved, ensuring efficient heat transfer and proper freezer function.
Coefficient of Performance
The Coefficient of Performance (COP) is an essential parameter for evaluating the efficiency of refrigerators and freezers. It is defined as the ratio of the cooling effect (heat removed from the cold reservoir) to the work input required to achieve that effect. A higher COP indicates a more efficient refrigeration cycle. The formula is given by:\[\text{COP} = \frac{Q_{cold}}{W}\]where \( Q_{cold} \) is the total heat removed from the freezer's interior, and \( W \) is the work input.

- **Understanding COP:** When you have a COP of 2.40, this means that for every unit of work energy supplied, the freezer can remove 2.40 units of heat. Essentially, a higher COP means better efficiency because the system can remove more heat per unit of energy expended.- **Calculating Work Done:** The work done by the freezer to remove the required heat is determined by rearranging the COP formula. By calculating the total heat removed in our exercise and knowing the COP value, one can easily compute the necessary work input to achieve the cooling effect.The COP is a critical factor when assessing the energy consumption and performance of cooling appliances, aligning with lower operational costs and improved sustainability.
Specific Heat Capacity
Specific heat capacity is a key concept in understanding how different substances absorb and retain heat. It represents the amount of heat per unit mass required to raise the temperature of a substance by one degree Celsius.\[Q = mc \Delta T\]where:- \( Q \) is the heat added,- \( m \) is the mass,- \( c \) is the specific heat capacity,- \( \Delta T \) is the change in temperature.

- **Different Substances, Different Capacities:** Water, for example, has a specific heat capacity of 4.186 kJ/kg°C, meaning it requires 4.186 kilojoules of energy to raise 1 kilogram by 1°C. Ice has a lower specific heat capacity of 2.09 kJ/kg°C, highlighting its different ability to change temperature with added or removed heat.- **Application in Thermodynamics:** In our exercise, we apply these specific heat capacities in sequential heat removal processes during the water cooling and ice formation. Different stages use the distinct heat capacities to accurately calculate energy removal at each temperature change.Understanding specific heat capacity allows us to predict and effectively manage temperature changes in various states of matter, critical for efficient thermal system designs.
Latent Heat
Latent heat is the energy absorbed or released by a substance during a phase change, such as from liquid to solid, without any change in temperature. This type of heat is crucial for understanding phase transitions. - **Latent Heat of Fusion:** In our example, when freezing water into ice, the latent heat involved, known as the heat of fusion, is absorbed. For water, this value is 334 kJ/kg, representing the energy needed per kilogram to transition from liquid to solid. - **Significance in Refrigeration:** Calculating the latent heat helps determine the total energy removal required when changing water to ice within a freezer. It's vital for accurately modeling how much heat needs to be extracted in operations involving multiple phase changes. Understanding latent heat is essential for designing efficient cooling systems and ensuring energy is utilized effectively during phase transitions. Whether heating, melting, or freezing, these transitions involve significant energy shifts critical to system performance.

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Most popular questions from this chapter

Consider a Diesel cycle that starts (at point \(a\) in Fig. 20.7\()\) with air at temperature \(T_{a}\) The air may be treated as an ideal gas. (a) If the temperature at point \(c\) is \(T_{c}\) , derive an expression for the efficiency of the cycle in terms of the compression ratio \(r .\) (b) What is the efficiency if \(T_{a}=300 \mathrm{K}, T_{c}=950 \mathrm{K}, \gamma=1.40\) and \(r=21.0 ?\)

A Carnot engine whose high-temperature reservoir is at 620 \(\mathrm{K}\) takes in 550 \(\mathrm{J}\) of heat at this temperature in each cycle and gives up 335 \(\mathrm{J}\) to the low-temperature reservoir. (a) How much mechanical work does the engine perform during each cycle? (b) What is the temperature of the low-temperature reservoir? (c) What is the thermal efficiency of the cycle?

A certain brand of freezer is advertised to use 730 \(\mathrm{kW} \cdot \mathrm{h}\) of energy per year. (a) Assuming the freezer operates for 5 hours each day, how much power does it require while operating? (b) If the freezer keeps its interior at a temperature of \(-5.0^{\circ} \mathrm{C}\) in a \(20.0^{\circ} \mathrm{C}\) room, what is its theoretical maximum performance coefficient? (c) What is the theoretical maximum amount of ice this freezer could make in an hour, starting with water at \(20.0^{\circ} \mathrm{C} ?\)

You are designing a Carnot engine that has 2 \(\mathrm{mol}\) of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 atm. With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work. (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 -kg block of ice originally at \(0.0^{\circ} \mathrm{C},\) using only the heat rejected by the engine?

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured-that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

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