Problem 12 - Pearson Physics

Question

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

Step-by-Step Solution

Answer

Rank is: D, B, A=C.

Step 1: Understanding Efficiency

The efficiency (\eta) of a heat engine is calculated as\[ \eta = \frac{Q_h - Q_c}{Q_h} \times 100 \% \]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 Efficiency for Engine A

For Engine A:- \(Q_h = 40 \text{ J}\)- \(Q_c = 20 \text{ J}\)The efficiency is:\[ \eta_A = \frac{40 - 20}{40} \times 100 \% = \frac{20}{40} \times 100 \% = 50\% \]

Step 3: Calculate Efficiency for Engine B

For Engine B:- \(Q_h = 140 \text{ J}\)- \(Q_c = 120 \text{ J}\)The efficiency is:\[ \eta_B = \frac{140 - 120}{140} \times 100 \% = \frac{20}{140} \times 100 \% \approx 14.29\% \]

Step 4: Calculate Efficiency for Engine C

For Engine C:- \(Q_h = 80 \text{ J}\)- \(Q_c = 40 \text{ J}\)The efficiency is:\[ \eta_C = \frac{80 - 40}{80} \times 100 \% = \frac{40}{80} \times 100 \% = 50\% \]

Step 5: Calculate Efficiency for Engine D

For Engine D:- \(Q_h = 240 \text{ J}\)- \(Q_c = 220 \text{ J}\)The efficiency is:\[ \eta_D = \frac{240 - 220}{240} \times 100 \% = \frac{20}{240} \times 100 \% \approx 8.33\% \]

Step 6: Rank Efficiencies

Now compare the efficiencies:- \(\eta_A = 50\%\)- \(\eta_C = 50\%\)- \(\eta_B \approx 14.29\%\)- \(\eta_D \approx 8.33\%\)Rank in increasing order: D (8.33%), B (14.29%), A (50%) = C (50%).

Key Concepts

Thermodynamics

Thermodynamics is all about the study of energy, especially how it moves and changes. It covers concepts like heat, work, and energy transfer. Think of it as a way to understand how different systems exchange energy. In our context, the focus is on heat engines. These are devices that convert heat energy into mechanical work. A common example is a car engine, where fuel combustion generates heat and is used to produce motion. What's intriguing about thermodynamics is how it lays down rules about energy. These rules, or laws, help us understand how energy can change forms but cannot be created or destroyed. They also tell us that a certain amount of energy will always be 'lost' as useless waste, usually as heat. These principles are key to analyzing the efficiency of engines, which we'll calculate using numbers from each engine's heat exchange.

Heat Absorption

In the context of a heat engine, heat absorption refers to the process where the engine takes in energy from a hot source. This heat is necessary for the engine to do work. The more heat an engine absorbs, the more potential it has to convert that energy into work.A crucial component of understanding heat engines is identifying how much of the absorbed heat can effectively be converted into work. In our exercise, heat absorption is represented by the variable \(Q_h\), which denotes the heat taken from the hot reservoir. For each engine in our problem set:

• Engine A absorbs \(40 \, \text{J}\).
• Engine B absorbs \(140 \, \text{J}\).
• Engine C absorbs \(80 \, \text{J}\).
• Engine D absorbs \(240 \, \text{J}\).

Understanding these values helps us determine how much energy the engine has to work with initially.

Energy Transfer

Energy transfer in heat engines involves moving energy from a hot body to a cooler one, converting part of it into work. This process is central to how heat engines operate. During energy transfer, an engine must expel some heat to the environment. The expelled heat, represented by \(Q_c\), does not contribute to the work done by the engine.For the engines we are considering:

• Engine A expels \(20 \, \text{J}\) to the cold reservoir.
• Engine B expels \(120 \, \text{J}\).
• Engine C expels \(40 \, \text{J}\).
• Engine D expels \(220 \, \text{J}\).

This expulsion is necessary, as no engine can convert all absorbed energy into work due to the Second Law of Thermodynamics. Analyzing both the heat absorbed and expelled allows us to calculate each engine's efficiency.

Physics Problem Solving

Physics problem solving is all about applying theoretical concepts to practical problems. In the exercise, we need to calculate the efficiency of each engine using its absorbed and expelled heat values. The formula \[ \eta = \frac{Q_h - Q_c}{Q_h} \times 100 \% \] helps us accomplish this by comparing the net useful energy (\(Q_h - Q_c\)) with the total energy input (\(Q_h\)).Let's break down the calculation for each engine:

• Engine A: \( \eta_A = \frac{40 - 20}{40} \times 100 \% = 50\% \)
• Engine B: \( \eta_B = \frac{140 - 120}{140} \times 100 \% \approx 14.29\% \)
• Engine C: \( \eta_C = \frac{80 - 40}{80} \times 100 \% = 50\% \)
• Engine D: \( \eta_D = \frac{240 - 220}{240} \times 100 \% \approx 8.33\% \)

Using this problem-solving approach, we can rank the engines from least to most efficient, helping us understand which engine makes the best use of its absorbed heat energy.