Question

A heat pump has a coefficient of performance of 5.0. If the heat pump absorbs 40.0 cal of heat from the cold outdoors in each cycle, what is the amount of heat expelled to the warm indoors?

Short answer

Answer: The heat pump expels 50.0 cal of heat to the warm indoors in each cycle.

Step-by-step solution

Write down the formula for the coefficient of performance

The formula for the coefficient of performance of a heat pump is given by: COP = Q_out / W_in Where COP is the coefficient of performance, Q_out is the heat expelled to the warm indoors and W_in is the work input.

Express the work input in terms of the heat absorbed and expelled

The heat absorbed from the cold outdoors Q_in and the work input are used to produce the heat expelled to the warm indoors. Therefore, we can write the relationship between Q_in, W_in, and Q_out as: Q_out = Q_in + W_in

Substitute the formula for Q_out in terms of Q_in and W_in into the formula for COP

Substitute the expression for Q_out from the previous step into the formula for COP: COP = (Q_in + W_in) / W_in

Solve for W_in

Rearrange the equation in order to solve for the work input: W_in = Q_in / (COP - 1)

Plug in the given values of COP and Q_in

The given COP is 5.0 and the heat absorbed from the cold outdoors (Q_in) in each cycle is 40.0 cal. Plug these values into the equation for W_in: W_in = 40.0 cal / (5.0 - 1)

Calculate the work input (W_in)

Now calculate the work input: W_in = 40.0 cal / 4 = 10.0 cal

Calculate the heat expelled to the warm indoors (Q_out)

Use the relationship between Q_in, W_in, and Q_out to find the heat expelled to the warm indoors: Q_out = Q_in + W_in = 40.0 cal + 10.0 cal = 50.0 cal The heat pump expels 50.0 cal of heat to the warm indoors in each cycle.

Key concepts

Coefficient of Performance

The coefficient of performance (COP) is a fundamental concept used to evaluate the efficiency of heat pumps. It is essentially a ratio that compares how much heat energy is moved between two places to how much work is required to do so. In formulaic terms, it is represented as:

• COP = \( \frac{Q_{out}}{W_{in}} \) - where \( Q_{out} \) is the heat expelled and \( W_{in} \) is the work input.

In simple terms, if a heat pump has a high COP, it is more efficient, as it can transfer a greater amount of heat energy using less work input. For instance, a COP of 5.0 means that five units of heat energy are transferred for every one unit of energy used to power the pump. This concept allows us to easily compare different heat pumps' efficiencies and choose the best one for a specific application.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It provides a framework for understanding how heat pump systems function, explaining the flow of heat energy between different areas. Central to this is the First Law of Thermodynamics, also known as the Law of Energy Conservation, which states that energy cannot be created or destroyed, only transformed. For heat pumps, this means:

• The energy taken in (from the cold source) and the energy put in (work input) must equal the energy expelled (to the warm space).

In practice, this concept explains why we can use work input to transfer heat from a cold area to a warm one, seemingly against the natural direction heat flows. Understanding thermodynamics helps us calculate and predict the performance of heating systems accurately.

Heat Transfer

Heat transfer refers to the movement of thermal energy from one location to another. In the context of a heat pump, it is the transfer of heat energy from outdoors (or another heat source) to indoor spaces. Heat transfer can occur in three different ways: conduction, convection, and radiation:

• Conduction: The transfer of heat through a material, like the chassis of the heat pump.
• Convection: The movement of heat through fluids, such as the refrigerant in heat pump systems.
• Radiation: The transfer of heat in the form of electromagnetic waves, though less significant in heat pumps.

In heating systems, the efficient transfer of heat is critical for improved performance. This relates directly to the coefficient of performance and the amount of work required to achieve certain heating results.

Caloric Calculations

Caloric calculations involve determining the amounts of heat absorbed or released during the operation of a heat pump. In heat pumps, these calculations are necessary for establishing different aspects, such as the amount of work required or the heat energy displaced.Typically, caloric calculations establish the relationships between:

• The heat energy taken from the environment (\( Q_{in} \)),
• The work input (\( W_{in} \)), and
• The heat expelled to heated spaces (\( Q_{out} \)).

For example, in our exercise, we found:

• 40 calories of heat absorbed (\( Q_{in} \)) plus 10 calories of work input result in 50 calories of heat expelled (\( Q_{out} \)).

Exact caloric calculations ensure that all energy transfers adhere to thermodynamic principles accurately, enabling more precise predictions of a system's performance.