Problem 42 - University Physics with Modern Physics

Question

20.42. Heat Pump. A heat pump is a heat engine run in reverse. In winter it pumps heat from the cold air outside into the warmer air inside the building, maintaining the building at a comfortable temperature. In summer it pumps heat from the cooler air inside the building to the warmer air outside, acting as an air conditioner. (a) If the outside temperature in winter is \(-5.0^{\circ} \mathrm{C}\) and the inside temperature is \(17.0^{\circ} \mathrm{C}\) , how many joules of heat will the heat pump deliver to the inside for each joule of electrical energy used to run the unit, assuming an ideal Carnot cycle? ( b) Suppose you have the option of using electrical resistance heating rather than a heat pump. How much electrical energy would you need in order to deliver the same amount of heat to the inside of the house as in part (a)? Consider a Carnot heat pump delivering heat to the inside of a house to maintain it at \(68^{\circ} \mathrm{F}\) . Show that the beat pump delivers less heat for each joule of electrical energy used to operate the unit as the outside temperature decreases. Notice that this behavior is opposite to the dependence of the efficiency of a Carnot heat engine on the difference in the reservoir temperatures. Explain why this is so.

Step-by-Step Solution

Answer

The heat pump has a COP of 13.19; it delivers 13.19 joules of heat per joule of electricity. Electrical resistance heating requires 13.19 times more energy. The heat pump's efficiency decreases with colder outside temperatures.

Step 1: Convert Temperatures to Kelvin

To work with the Carnot cycle, temperatures must be in Kelvin. Convert the outside winter temperature and the inside temperature to Kelvin: \[ T_{outside} = -5.0\, ^{\circ}C + 273.15 = 268.15\, K \]\[ T_{inside} = 17.0\, ^{\circ}C + 273.15 = 290.15\, K \]

Step 2: Calculate the Coefficient of Performance (COP) for the Heat Pump

The Coefficient of Performance (COP) for a heat pump is calculated as follows:\[ \text{COP} = \frac{T_{inside}}{T_{inside} - T_{outside}} \]Substitute the Kelvin temperatures calculated:\[ \text{COP} = \frac{290.15}{290.15 - 268.15} = \frac{290.15}{22} \approx 13.19 \]

Step 3: Interpret the COP for Energy Transfer (a)

The COP indicates that for every joule of electrical energy used, the heat pump delivers approximately 13.19 joules of heat to the inside of the building.

Step 4: Compare with Electrical Resistance Heating (b)

Electrical resistance heating systems have a COP of 1, meaning 1 joule of electrical energy delivers 1 joule of heat. To deliver the same amount of heat as the heat pump, which delivers 13.19 joules of heat per joule of electricity, the electrical system would need 13.19 times more energy.

Step 5: Explain the Behavior of the Heat Pump Efficiency

As the outside temperature decreases, the COP decreases since the denominator \((T_{inside} - T_{outside})\) increases. This behavior is opposite to a Carnot engine because a heat engine's efficiency depends on maximizing the temperature difference, while a heat pump's performance improves with smaller temperature differences to move heat efficiently.

Key Concepts

Carnot Cycle

The Carnot cycle is a theoretical construct in thermodynamics. It represents the most efficient cycle that a heat engine can operate on. For a heat pump, which is a heat engine running in reverse, the Carnot cycle provides a benchmark for maximum theoretical efficiency. This cycle consists of two isothermal processes and two adiabatic processes. Heat pumps work by moving heat from a cooler area to a warmer area, which naturally goes against what's usual, so they require external energy.

• Isothermal processes involve heat transfer while maintaining a constant temperature.
• Adiabatic processes involve changes in pressure and volume but with no heat transfer.

Though idealized, understanding the Carnot cycle helps us grasp the limits of real-world performance. It shows how efficiency is dependent on temperature differences between the inside and outside.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is essential for understanding the efficiency of heat pumps. Unlike efficiency in engines, which is the ratio of work output to heat input, COP measures a heat pump's effectiveness in transferring heat compared to the energy it requires. It is defined by the formula:\[\text{COP} = \frac{T_{\text{inside}}}{T_{\text{inside}} - T_{\text{outside}}}\]COP tells us how much heat is moved per unit of energy consumed by the pump. Higher COP implies greater efficiency. For the heat pump scenario described initially, this COP of 13.19 means that for every joule of electrical energy used, approximately 13.19 joules of heat are transferred inside. Factors such as temperature differences greatly impact the COP, showing less performance as exterior temperatures drop.

Electrical Resistance Heating

Electrical resistance heating is a straightforward way to generate heat. In this process, electrical energy is directly transformed into heat via resistors. This method has a constant COP of 1, meaning every joule of electricity contributes one joule of heat. While simple and effective, electrical resistance heating lacks the efficiency of a heat pump. In comparison, a heat pump, with its ability to move existing heat rather than directly create it, typically offers far superior efficiency. For example, under the conditions described, a heat pump can deliver 13.19 joules of heat for each joule of electricity, proving much more energy-efficient than electrical resistance methods.

Thermodynamics

Thermodynamics is the science of energy transformations. It helps us understand how heat and work interact in various systems. Heat pumps operate on principles of thermodynamics, particularly the laws governing energy conservation and the direction of heat flow. Key concepts include:

• The First Law of Thermodynamics, which asserts that energy cannot be created or destroyed, only transformed.
• The Second Law, highlighting that heat naturally flows from hot to cold areas unless energy is input to reverse that flow.

These laws underlie how heat pumps can move heat against its natural direction with the help of external work, exemplifying the clever manipulation of energy flows to achieve desired heating or cooling in homes.

Temperature Conversion

Temperature conversion is vital when using thermodynamic equations, as computations often require using Kelvin. The Kelvin scale is an absolute temperature scale, starting at absolute zero.To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature.For example, converting the temperatures involved in our initial exercise:

• Outside temperature: \[-5.0^{\circ}C = 268.15\, K\]
• Inside temperature: \[17.0^{\circ}C = 290.15\, K\]

Converting to Kelvin is indispensable as it normalizes calculations, ensuring consistency across physics and thermodynamics problems.