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The electric motor of a heat pump transfers energy as heat from the outdoors, which is at \(-5.0^{\circ} \mathrm{C},\) to a room that is at \(17^{\circ} \mathrm{C}\). If the heat pump were a Carnot heat pump (a Carnot engine working in reverse), how much energy would be transferred as heat to the room for each joule of electric energy consumed?

Short Answer

Expert verified
13.2 joules of energy is transferred as heat for each joule of electric energy consumed.

Step by step solution

01

Convert Temperatures to Kelvin

The first step is to convert the temperatures from Celsius to Kelvin, as the Carnot efficiency formula requires absolute temperatures. For the outdoor temperature, \(-5.0^{\circ} \mathrm{C}\), we convert it by adding 273.15: \[ T_c = -5.0 + 273.15 = 268.15\, \mathrm{K} \]For the indoor temperature, \(17^{\circ} \mathrm{C}\), we convert it by adding 273.15: \[ T_h = 17 + 273.15 = 290.15 \, \mathrm{K} \]
02

Use the Carnot Efficiency Formula for Heat Pump

The efficiency of a Carnot heat pump is determined by:\[ \text{COP} = \frac{T_h}{T_h - T_c} \]Substitute the temperatures we calculated in Step 1:\[ \text{COP} = \frac{290.15}{290.15 - 268.15} = \frac{290.15}{22} \approx 13.188 \]
03

Calculate Energy Transferred as Heat

In this context, the coefficient of performance (COP) indicates how much energy is transferred as heat to the room for each joule of electric energy consumed. With a COP of 13.188, for each joule of electrical energy consumed, the energy transferred as heat is 13.188 joules.

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Key Concepts

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

Carnot efficiency
The concept of Carnot efficiency is a fundamental principle in thermodynamics, but rather than focusing on efficiency in terms of work output, the efficiency of a heat pump refers to how well it can use input energy to transfer heat.
A Carnot heat pump operates on the reverse Carnot cycle and is considered the ideal model for heat pumps. It assumes that energy is moved with maximum efficiency between two heat reservoirs at different temperatures. However, for heat pumps, we often discuss this efficiency in terms of the Coefficient of Performance (COP), not as a percentage of energy converted into work. This is because heat pumps move heat rather than produce mechanical work.
  • The efficiency depends on the temperature difference between the heat source and sink.
  • Since a Carnot heat pump is an idealized model, real-world heat pumps will usually have a lower efficiency than predicted by Carnot's theoretical limits.
  • The Carnot efficiency provides a benchmark to compare the performance of real heat pumps.
Coefficient of performance (COP)
The Coefficient of Performance (COP) is a key measure for understanding how effectively a heat pump operates. Unlike efficiency, which is often expressed as a percentage, COP is a ratio that compares the heat output to the energy input.
COP indicates how much heat energy is moved to the room for each unit of electrical energy consumed. In our exercise, this was calculated using the temperatures converted to Kelvin.
  • The formula to calculate the COP of a Carnot heat pump is: \[\text{COP} = \frac{T_h}{T_h - T_c}\], where \(T_h\) is the temperature of the room and \(T_c\) is the outdoor temperature.
  • A higher COP means that the heat pump is more effective, moving more heat energy per unit of electrical energy.
  • Carnot provides the theoretical maximum COP, which is difficult to achieve in practical applications due to various losses.
Understanding COP helps in selecting an efficient system and determining potential energy savings.
Energy transfer in heat pumps
Energy transfer in heat pumps involves moving heat from a cold area to a warm area, effectively operating as a reverse refrigerator. In the context of our exercise, the heat pump transfers heat from the cold outdoors to the warmer indoors.
This transfer of heat is not merely heating up the space but rather shifting energy from one environment to another through the use of a working fluid cycle.
  • In a Carnot heat pump, this transfer is maximized under ideal conditions, as suggested by the high COP value.
  • For each joule of electricity, a Carnot heat pump can ideally transfer multiple joules of heat energy, depending on the COP. In our example, it could transfer 13.188 joules of heat for every joule of energy consumed.
  • Heat pumps utilize a cycle of condensation and evaporation, where a refrigerant absorbs heat from one area and releases it in another.
By understanding this energy transfer process, users can appreciate why heat pumps are a more efficient option compared to conventional heating methods that directly convert electricity to heat.

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Most popular questions from this chapter

To make ice, a freezer that is a reverse Carnot engine extracts \(42 \mathrm{~kJ}\) as heat at \(-15^{\circ} \mathrm{C}\) during each cycle, with coefficient of performance \(5.7 .\) The room temperature is \(30.3^{\circ} \mathrm{C}\). How much (a) energy per cycle is delivered as heat to the room and (b) work per cycle is required to run the freezer?

An ideal gas undergoes a reversible isothermal expansion at \(77.0^{\circ} \mathrm{C},\) increasing its volume from \(1.30 \mathrm{~L}\) to \(3.40 \mathrm{~L}\). The entropy change of the gas is \(22.0 \mathrm{~J} / \mathrm{K}\). How many moles of gas are present?

Suppose \(0.550 \mathrm{~mol}\) of an ideal gas is isothermally and reversibly expanded in the four situations given below. What is the change in the entropy of the gas for each situation? $$\begin{array}{lcccc}\hline \text { Situation } & (\mathrm{a}) & (\mathrm{b}) & (\mathrm{c}) & (\mathrm{d}) \\\\\hline \text { Temperature }(\mathrm{K}) & 250 & 350 & 400 & 450 \\\\\text { Initial volume }\left(\mathrm{cm}^{3}\right) & 0.200 & 0.200 & 0.300 & 0.300 \\\\\text { Final volume }\left(\mathrm{cm}^{3}\right) & 0.800 & 0.800 & 1.20 & 1.20 \\\\\hline\end{array}$$

The motor in a refrigerator has a power of \(200 \mathrm{~W}\). If the freezing compartment is at \(270 \mathrm{~K}\) and the outside air is at \(300 \mathrm{~K},\) and assuming the efficiency of a Carnot refrigerator, what is the maximum amount of energy that can be extracted as heat from the freezing compartment in \(10.0 \mathrm{~min} ?\)

Expand \(1.00 \mathrm{~mol}\) of an monatomic gas initially at \(5.00 \mathrm{kPa}\) and \(600 \mathrm{~K}\) from initial volume \(V_{i}=1.00 \mathrm{~m}^{3}\) to final volume \(V_{f}=2.00 \mathrm{~m}^{3}\). At any instant during the expansion, the pressure \(p\) and volume \(V\) of the gas are related by \(p=5.00 \exp \left[\left(V_{i}-V\right) / a\right],\) with \(p\) in kilopascals, \(V_{i}\) and \(V\) in cubic meters, and \(a=1.00 \mathrm{~m}^{3} .\) What are the final (a) pressure and (b) temperature of the gas? (c) How much work is done by the gas during the expansion? (d) What is \(\Delta S\) for the expansion? (Hint: Use two simple reversible processes to find \(\Delta S .)\)

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