Problem 12 - Pearson Physics

Question

Rank Determine the efficiencies of the engines \(\mathrm{A}\) through D, described below, and then rank them in order of increasing efficiency. Indicate ties where appropriate. \begin{tabular}{|c|c|c|c|c|} \hline Engine & A & B & C & D \\ \hline \(\mathbf{Q}_{\mathbf{h}}(\mathbf{J})\) & 40 & 140 & 80 & 240 \\ \hline \(\mathbf{Q}_{\mathbf{c}}(\mathbf{J})\) & 20 & 120 & 40 & 220 \\ \hline \end{tabular}

Step-by-Step Solution

Answer

D, B, A=C

Step 1: Determine the Efficiency Formula

The efficiency of a heat engine is calculated using the formula: \( \eta = 1 - \frac{Q_c}{Q_h} \), where \( Q_h \) is the heat absorbed from the hot reservoir and \( Q_c \) is the heat expelled to the cold reservoir.

Step 2: Calculate the Efficiency of Engine A

For Engine A, \( Q_h = 40 \) J and \( Q_c = 20 \) J. Substituting into the efficiency formula: \[ \eta_A = 1 - \frac{20}{40} = 1 - 0.5 = 0.5 \text{ or } 50\% \].

Step 3: Calculate the Efficiency of Engine B

For Engine B, \( Q_h = 140 \) J and \( Q_c = 120 \) J. Substituting into the efficiency formula: \[ \eta_B = 1 - \frac{120}{140} = 1 - \frac{6}{7} = \frac{1}{7} \approx 0.1429 \text{ or } 14.29\% \].

Step 4: Calculate the Efficiency of Engine C

For Engine C, \( Q_h = 80 \) J and \( Q_c = 40 \) J. Substituting into the efficiency formula: \[ \eta_C = 1 - \frac{40}{80} = 1 - 0.5 = 0.5 \text{ or } 50\% \].

Step 5: Calculate the Efficiency of Engine D

For Engine D, \( Q_h = 240 \) J and \( Q_c = 220 \) J. Substituting into the efficiency formula: \[ \eta_D = 1 - \frac{220}{240} = 1 - \frac{11}{12} = \frac{1}{12} \approx 0.0833 \text{ or } 8.33\% \].

Step 6: Rank the Engines by Efficiency

Comparing efficiencies: \( \eta_A = 50\% \), \( \eta_C = 50\% \), \( \eta_B = 14.29\% \), and \( \eta_D = 8.33\% \). Therefore, the order of increasing efficiency is: D, B, A=C.

Key Concepts

Thermal Efficiency

When we talk about thermal efficiency in the context of heat engines, we're addressing the relationship between the energy input and the useful work output. It's a measure of how well an engine converts the heat from its fuel into work. Mathematically, thermal efficiency \( \eta \) is given by:\[ \eta = 1 - \frac{Q_c}{Q_h} \]where:

• \( Q_h \) is the heat absorbed from the hot reservoir (input energy).
• \( Q_c \) is the heat expelled to the cold reservoir (waste energy).

Thermal efficiency is expressed as a percentage. The closer the efficiency is to 1 (or 100%), the better the engine is at converting heat into work. Understanding this concept helps us compare the effectiveness of different engines, just like determining the efficiency of Engines A through D in the exercise.

Heat Transfer

Heat transfer is an essential physical process in heat engines where energy is transferred from the hot reservoir to the engine, and then from the engine to the cold reservoir. This transfer occurs due to temperature differences. Heat naturally flows from a higher temperature to a lower temperature region.

• In the context of our exercise, \( Q_h \) represents the heat transferred from the hot source to the engine. It is energy intake.
• \( Q_c \) is the heat that the engine rejects to the cold reservoir. This is often the fraction of energy that cannot be converted to work.

Effective heat transfer is crucial for a heat engine's performance. A larger difference between \( Q_h \) and \( Q_c \) typically leads to higher efficiency because less energy is lost. An engine with low \( Q_c \) has less waste heat, implying better performance.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat and temperature and their relation to other forms of energy. The principles of thermodynamics underpin the functioning of all heat engines. The most relevant principle for heat engines is the Second Law of Thermodynamics, which states that in any energy exchange, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state, i.e., there will arise entropy — randomness or disorder.

• In simple terms, it explains why we can't have a perfect engine that converts all heat into work. Some energy is always wasted, increasing entropy.
• Heat engines operate on a cycle, transforming some of the input heat into work and expelling the rest, hence balancing the increase in entropy.

By understanding thermodynamics, students can better grasp why no engine can be 100% efficient, which is vital for analyses similar to the one performed on engines A through D.