Question

In an effort to conserve energy in a heat-engine cycle, somebody suggests incorporating a refrigerator that will absorb some of the waste energy \(Q_{L}\) and transfer it to the energy source of the heat engine. Is this a smart idea? Explain.

Short answer

Explain. Answer: No, incorporating a refrigerator to transfer waste energy back to the heat source is not a smart idea for conserving energy. This is because, due to the second law of thermodynamics, no heat engine or refrigerator can have 100% efficiency or an infinite COP. The energy used by the refrigerator to transfer waste heat will always be more than the extra work obtained by the heat engine, leading to additional energy waste instead of conservation.

Step-by-step solution

Consider the heat engine

A heat engine's purpose is to convert heat energy from a high-temperature source (usually denoted as \(Q_{H}\)) into useful work \(W\). During this conversion, a part of the heat energy is lost as waste energy to a colder reservoir (denoted as \(Q_{L}\)). The efficiency of a heat engine is given by: \[ \eta = \frac{W}{Q_{H}} = 1 - \frac{Q_{L}}{Q_{H}} \]

Consider a refrigerator

A refrigerator's purpose is to transfer heat from a low-temperature source (\(Q_{cold}\)) to a high-temperature source (\(Q_{hot}\)), by consuming work \(W_{R}\). The coefficient of performance (COP) of a refrigerator can be given by: \[ COP = \frac{Q_{cold}}{W_{R}} \]

Mutual relation between heat engine and refrigerator

By incorporating a refrigerator to transfer the waste energy \(Q_{L}\) back to the heat source, we are aiming to utilize \(Q_{L}\) and save energy. The refrigerator will transfer \(Q_{L}\) by consuming work \(W_{R}\). Given the COP of the refrigerator, we can calculate the work needed for the refrigerator to transfer \(Q_{L}\): \[W_{R} = \frac{Q_{L}}{COP}\]

Determine if it is a smart idea

Since the heat engine produces work \(W\) and the refrigerator consumes work \(W_{R}\) to transfer heat, the net work done by the system can be expressed as: \[W_{net} = W - W_{R}\] Substitute the efficiency of the heat engine and the COP of the refrigerator in the net work done formula: \[W_{net} = \eta Q_{H} - \frac{Q_{L}}{COP}\] Recall that in our case, \(Q_{L}\) is being transferred back to the heat source. Therefore: \[W_{net} = \eta Q_{H} - \frac{\eta Q_{H}}{COP}\] The system conserves energy and is considered a smart idea if \(W_{net}\) is greater than the work done without the refrigerator. However, due to the second law of thermodynamics, no heat engine or refrigerator can have 100% efficiency or an infinite COP. The energy used by the refrigerator to transfer waste heat will always be more than the extra work obtained by the heat engine. This energy will be wasted in the form of heat or other inefficiencies. Therefore, incorporating a refrigerator to transfer waste energy back to the heat source will not lead to energy conservation, and thus, is not a smart idea.