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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

Expert verified
(a) 393,858 J (b) 164,108 J (c) 557,966 J

Step by step solution

01

Calculate Heat to Cool Water to 0°C

To start, calculate the amount of heat required to lower the temperature of 1.80 kg of water from 25.0°C to 0°C using the specific heat capacity of water, which is 4186 J/kg°C:\[ q_1 = m c \Delta T = 1.80 \times 4186 \times (0 - 25.0) \]\[ q_1 = -1.80 \times 4186 \times 25.0 = -188,370 \text{ J} \]
02

Calculate Heat to Freeze Water at 0°C

Next, calculate the heat required to freeze the water at 0°C. This requires the latent heat of fusion for water, 334,000 J/kg:\[ q_2 = m L_f = 1.80 \times 334,000 \]\[ q_2 = 601,200 \text{ J} \]
03

Calculate Heat to Cool Ice to -5°C

Then, calculate the heat required to cool the ice from 0°C to -5.0°C using the specific heat capacity of ice, 2108 J/kg°C:\[ q_3 = m c \Delta T = 1.80 \times 2108 \times (-5.0) \]\[ q_3 = -18,972 \text{ J} \]
04

Summing Total Heat Removed

The total amount of heat \( Q \) removed is the sum of all three calculated values:\[ Q = q_1 + q_2 + q_3 = -188,370 + 601,200 + (-18,972) \]\[ Q = 393,858 \text{ J} \]
05

Calculate Electrical Energy Consumed

The coefficient of performance (COP) is given as 2.40. The equation for COP in terms of heat removed \( Q \) and work \( W \) is:\[ \text{COP} = \frac{Q}{W} \Longrightarrow W = \frac{Q}{\text{COP}} \]\[ W = \frac{393,858}{2.40} \]\[ W \approx 164,108 \text{ J} \]
06

Calculate Wasted Heat Delivered

The wasted heat \( Q' \) delivered to the room is the sum of the work done and the heat extracted:\[ Q' = Q + W = 393,858 + 164,108 \]\[ Q' = 557,966 \text{ J} \]

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Key Concepts

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

Heat Transfer
In this context, heat transfer refers to the process of energy moving from one substance to another. Here, it is crucial in changing the water from a liquid at 25°C to ice at -5°C. This process involves three main phases: cooling the water to 0°C, freezing it at 0°C, and then cooling the ice further to -5°C.
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
The coefficient of performance (COP) is a measure of efficiency for refrigerators and freezers. It is defined as the ratio of the heat removed from the freezer to the work input.
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
Latent heat refers to the energy absorbed or released during a phase change of a substance, without a change in temperature.
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.
The energy value is critical for accurate transition estimation from one phase to another, especially in thermal systems like freezers.
Specific Heat Capacity
Specific heat capacity is the amount of heat energy required to raise the temperature of a unit of mass of a substance by one degree Celsius.
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.
This concept is essential in defining how substances absorb and release heat energy in transitional temperature intervals, thus playing a crucial role in thermal management systems.

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Most popular questions from this chapter

For a refrigerator or air conditioner, the coefficient of performance \(K\) (often denoted as COP) is, as in Eq. (20.9), the ratio of cooling output \(Q_C\) 0 to the required electrical energy input \(W\) , both in joules. The coefficient of performance is also expressed as a ratio of powers, $$K = {(Q_C ) /t \over (W) /t}$$ where \(Q_C /t\) is the cooling power and \(W /t\) is the electrical power input to the device, both in watts. The energy efficiency ratio (\(EER\)) is the same quantity expressed in units of Btu for \(Q_C\) and \(W \cdot h\) for \(W\) . (a) Derive a general relationship that expresses \(EER\) in terms of \(K\). (b) For a home air conditioner, \(EER\) is generally determined for a 95\(^\circ\)F outside temperature and an 80\(^\circ\)F return air temperature. Calculate \(EER\) for a Carnot device that operates between 95\(^\circ\)F and 80\(^\circ\)F. (c) You have an air conditioner with an \(EER\) of 10.9. Your home on average requires a total cooling output of \(Q_C = 1.9 \times 10^{10} J\) per year. If electricity costs you 15.3 cents per \(kW \cdot h\), how much do you spend per year, on average, to operate your air conditioner? (Assume that the unit's \(EER\) accurately represents the operation of your air conditioner. A \(seasonal\) \(energy\) \(efficiency\) \(ratio\) (\(SEER\)) is often used. The \(SEER\) is calculated over a range of outside temperatures to get a more accurate seasonal average.) (d) You are considering replacing your air conditioner with a more efficient one with an \(EER\) of 14.6. Based on the \(EER\), how much would that save you on electricity costs in an average year?

Compare the entropy change of the warmer water to that of the colder water during one cycle of the heat engine, assuming an ideal Carnot cycle. (a) The entropy does not change during one cycle in either case. (b) The entropy of both increases, but the entropy of the colder water increases by more because its initial temperature is lower. (c) The entropy of the warmer water decreases by more than the entropy of the colder water increases, because some of the heat removed from the warmer water goes to the work done by the engine. (d) The entropy of the warmer water decreases by the same amount that the entropy of the colder water increases.

A Carnot engine is operated between two heat reservoirs at temperatures of 520 \(K\) and 300 \(K\). (a) If the engine receives 6.45 \(kJ\) of heat energy from the reservoir at 520 \(K\) in each cycle, how many joules per cycle does it discard to the reservoir at 300 \(K\)? (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?

A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (a) If in each cycle the refrigerator receives 415 J of heat energy from the reservoir at 270 K, how many joules of heat energy does it deliver to the reservoir at 320 K? (b) If the refrigerator completes 165 cycles each minute, what power input is required to operate it? (c) What is the coefficient of performance of the refrigerator?

As a budding mechanical engineer, you are called upon to design a Carnot engine that has 2.00 mol of a monatomic ideal gas as its working substance and operates from a high temperature reservoir at 500\(^\circ\)C. The engine is to lift a 15.0-kg weight 2.00 m per cycle, using 500 J of heat input. The gas in the engine chamber can have a minimum volume of 5.00 \(L\) during the cycle. (a) Draw a \(pV\)-diagram for this cycle. Show in your diagram where heat enters and leaves the gas. (b) What must be the temperature of the cold reservoir? (c) What is the thermal efficiency of the engine? (d) How much heat energy does this engine waste per cycle? (e) What is the maximum pressure that the gas chamber will have to withstand?

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